7.4: Solving Trigonometric Equations Using Identities

Suggested Prerequisite Topics

This section is designed assuming you understand the following topics from Algebra.

• Solving radical equations
• Solving logarithmic equations (to discuss extraneous solutions)

Learning Objectives

• Use an identity to solve a trigonometric equation.
• Rewrite a sum or difference of sine and cosine as a single sinusoidal function.
• Use technology to approximate solutions to equations involving trigonometric functions.

Having a solid foundation of how trigonometric equations are solved, we now integrate our knowledge of trigonometric identities to further expand our equation-solving abilities.

Solving Equations Involving Different Trigonometric Functions Having the Same Arguments

The equations encountered in the previous section were ideal in that each featured only one trigonometric function, and all occurrences of that function within the equation shared the same argument. That is, we only encountered equations like the following:\[ \begin{array}{rcl}
\sin\left( \theta \right) & = & -0.6428 \\
\tan\left( 2x \right) & = & \sqrt{3} \\
4 \tan^2\left( \theta \right) + 3 & = & 15 \\
2\cos^2\left( \theta \right) - \cos\left( \theta \right) - 1 & = & 0 \\
\end{array} \nonumber \]Each of these equations features only one trigonometric function. In the fourth equation, that trigonometric function (the cosine) occurs twice, but in both cases, the arguments are identical (\( \theta \)).

How would we solve a trigonometric equation if it contains two different trigonometric functions? For example, if we were tasked with solving\[\sec^{2}(x) = \tan(x) + 3,\nonumber\]what would we do?

Ultimately, the answer is that we use a mixture of the trigonometric identities learned in this course and algebraic manipulations learned in Algebra. The worst thing I could do at this point is throw a bunch of examples at you showing how to solve specific equations - that approach will make the single equation-solving tactic seem like a dozen different tactics. To avoid this, let's start by reviewing how we solved trigonometric equations in the previous section and codify this into a tactic.

When we were given an equation of the form\[ A \, \mathrm{trig}\left( Bx + C \right) + D = E, \nonumber \]we first isolated the trigonometric function (using techniques from Algebra) to a form like\[ \mathrm{trig}\left( Bx + C \right) = c, \nonumber \]where \( c \) is a real number. I also suggested a "substitution" technique if the argument of the trigonometric function wasn't a simple variable like \( \theta \) or \( x \). Let's suppose we use that technique to let \( u = Bx + C \). Then our equation would simplify down to\[ \mathrm{trig}\left( u \right) = c. \nonumber \]At this point, we used the sign of \( c \) to help us determine the quadrants and reference angle associated with our solution. Once those two pieces of information were found, we stated the answers.

This review leads to the following tactic for solving "basic" trigonometric equations.

Strategy for Solving "Basic" Trigonometric Equations

Given any trigonometric equation of the form\[ \mathrm{trig}(u) = c, \label{trigeqn}\]where "\(\mathrm{trig}\)" could be \( \sin, \cos, \tan, \csc, \sec, \) or \( \cot \), do the following:

1. If the value of \( c \) is such that \( u \) is a quadrantal angle, skip to Step 5.
2. Use the sign of \( c \) to determine in which quadrants \( u \) terminates.
3. Find the reference angle, \( \hat{u} \).
4. Using the reference angle from Step 3, write the solutions associated with the quadrants listed in Step 2. If possible, compress the solution set to a single equation (this is most often the case with equations involving the tangent or the cotangent functions).
5. If \( u \) is a function of another variable, solve the equations in Step 4 (or Step 1, if the angle is a quadrantal angle) for the "true" variable.
6. Check for extraneous solutions!

Our ultimate goal when solving any trigonometric equation is to get it into the form of Equation \ref{trigeqn}.

Example \( \PageIndex{1} \)

Solve the following equations and list the solutions which lie in the interval \([0,2\pi)\).

1. \(3\cos(x) \sin^{2}(x) = \sin^{2}(x)\)
2. \(\sec^{2}(x) = \tan(x) + 3\)
3. \(\cos ^2 \left(A\right)-\sin ^2 \left(A\right)-\sin \left(A\right)=0\)

Solutions

1. We resist the temptation to divide both sides of \(3\cos(x)\sin^{2}(x) = \sin^{2}(x)\) by \(\sin^{2}(x)\).¹ Instead, we look at our desired destination - Equation \ref{trigeqn}. One thing is certain about that desired form: We need all trigonometric functions on one side. Let's start there.\[ \begin{array}{rrrclcl}
   & & 3\cos(x) \sin^2(x) & = & \sin^2(x) & & \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 3\cos(x) \sin^2(x) - \sin^2(x) & = & 0 & \quad & \left( \text{move all variable expressions to one side} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \sin^{2}(x) \left[3 \cos(x) - 1\right] & = & 0 & \quad & \left( \text{factor out the GCF}\right) \\
   \end{array}\nonumber\]We now use the Zero Factor Property from Algebra to break the equation into two different (and simpler) equations.\[\begin{array}{rrcccc}
   & \sin^{2}(x) = 0 & \quad & \text{or} & \quad & 3\cos(x) - 1 = 0 \\
   \\
   \implies & \sin\left( x \right) = 0 & \quad & \text{or} & \quad & \cos\left( x \right) = \dfrac{1}{3} \\
   \end{array} \nonumber\]These equations match the form of Equation \ref{trigeqn}, so solving them will mirror what we did in the previous section.

   For \( \sin(x) = 0 \)

   The right side of \( \sin(x) = 0 \) implies a quadrantal angle.

   

   From the graph of \( y = \sin\left( x \right)\), we know the solution to this equation is \(x = \pi k\), with \(x = 0\) and \(x = \pi\) being the two solutions which lie in \([0,2\pi)\).

   For \( \cos(x) = \frac{1}{3} \)

   Since the cosine is positive, its argument is an angle that terminates in \( \mathrm{QI} \) or \( \mathrm{QIV} \). Additionally, the ratio, \(\frac{1}{3}\), is not a special ratio, so we will eventually need the use of technology. For now, we note that the reference angle is \(\hat{x} = \cos^{-1}\left(\left|\frac{1}{3}\right|\right) = \cos^{-1}\left(\frac{1}{3}\right)\). Hence, the two families of solutions are\[ \begin{array}{rrcl}
   \text{Solutions associated with QI}: & x & = & \cos^{-1}\left( \dfrac{1}{3} \right) + 2 \pi k \\
   \\
   \text{Solutions associated with QIV}: & x & = & 2 \pi - \cos^{-1}\left( \dfrac{1}{3} \right) + 2 \pi k \\
   \end{array} \nonumber \]where \( k \in \mathbb{Z} \). We find the two solutions which lie in \([0,2\pi)\) to be\[x = \cos^{-1}\left(\dfrac{1}{3}\right) \quad \text{and} \quad x = 2\pi - \cos^{-1}\left(\dfrac{1}{3}\right).\nonumber \]
   Thus, the solutions to the original equation on the interval \( [0, 2 \pi) \) are\[ x = 0, \, x = \cos^{-1}\left(\dfrac{1}{3}\right), \, x = \pi, \, \text{and} \, x = 2\pi - \cos^{-1}\left(\frac{1}{3}\right). \nonumber \]Grabbing a calculator (and rounding to four decimal places), we get\[ x = 0, \, x = \pi, \, x \approx 1.2310, \, \text{and} \, x \approx 5.0522.  \nonumber \]
2. Again, our ultimate goal is to get an equation (or equations) equivalent to Equation \ref{trigeqn}. At the very least, this means we need to move all the trigonometric functions to one side of the equation.\[ \begin{array}{rrrclcl}
   & & \sec^2(x) & = & \tan(x) + 3 & & \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \sec^2(x) - \tan(x) & = & 3 & \quad & \left( \text{move all variable expressions to one side} \right)\\
   \end{array} \nonumber \]Again, we have a mixture of different trigonometric functions on the left side, but this time, Algebra (specifically, factoring) is not going to help (do you see why?). Instead, we look for any trigonometric identities that might be of use. The \( \sec^2(x) \) in an equation with \( \tan(x) \) screams "Pythagorean Identity!"\[\begin{array}{rrrclcl}
   & & \sec^{2}(x) - \tan(x) & = & 3 & & \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & \tan^{2}(x) + 1 - \tan(x) & = & 3 & \quad & \left( \text{Pythagorean Identities} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \tan^{2}(x) - \tan(x) -2 & = & 0 & \quad & \left( \text{subtract}\, 3 \, \text{from both sides} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & u^2 - u -2 & = & 0 & \quad & \left( \text{substitute }u = \tan(x) \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \left( u + 1 \right)\left( u - 2 \right) & = & 0 & \quad & \left( \text{factor} \right) \\
   \end{array}\nonumber\]This gives \(u = -1\) or \(u = 2\). Since \(u = \tan(x)\), we have \(\tan(x) = -1\) or \(\tan(x) = 2\).

   For \( \tan(x) = -1 \)

   Since the tangent is negative, we know the argument terminates in either \( \mathrm{QII} \) or \( \mathrm{QIV} \). Moreover, the reference angle is \( \frac{\pi}{4} \). Hence,\[ \begin{array}{rrcl}
   \text{Solutions associated with QII and IV}: & x & = & \dfrac{3\pi}{4} + \pi k \\
   \end{array} \nonumber \]where \( k \in \mathbb{Z} \). Notice that we used the fact that the period of the tangent is \( \pi \) to compress our two solution sets into one. Letting \( k = 0 \) and then \( k = 1 \), we get the solutions \( x = \frac{3 \pi}{4} \) and \( x = \frac{7 \pi}{4} \), respectively. Any further, and our values for \( x \) will go beyond the requested interval.

   For \( \tan(x) = 2 \)

   The tangent is positive in \( \mathrm{QI} \) and \( \mathrm{QIII} \). We should recognize that the right side is not a special ratio for the tangent function. Therefore, we set the reference angle to \( \hat{x} = \tan^{-1}(|2|) = \tan^{-1}(2) \). Thus,\[ \begin{array}{rrcl}
   \text{Solutions associated with QI and III}: & x & = & \tan^{-1}(2) + \pi k \\
   \end{array} \nonumber \]where \( k \in \mathbb{Z} \). On \( \left[ 0, 2\pi \right) \), \( x = \tan^{-1}(2) \) and \( x = \pi + \tan^{-1}(2)\) are the solutions.
   
   Hence, the solution set to the original equation on the interval \( \left[ 0, 2 \pi \right) \) is\[ x \in \left\{ \tan^{-1}(2), \, \dfrac{3 \pi}{4}, \, \pi + \tan^{-1}(2), \, \dfrac{7 \pi}{4} \right\}, \nonumber \]where \( \tan^{-1}(2) \approx 1.1071 \) and \( \pi + \tan^{-1}(2) \approx 4.2487 \).
3. Since this equation involves both the sine and cosine functions, we will need to either perform some Algebra or look for trigonometric identities to get to a point where we have a simple equation (or equations) like \ref{trigeqn}. Factoring out the GCF is out of the question because the three terms on the left side of\[ \cos ^2 \left(A\right)-\sin ^2 \left(A\right)-\sin \left(A\right)=0 \nonumber \]do not share a common factor. Therefore, we immediately look for obvious trigonometric identities. We know by this point that any time we encounter the square of a trigonometric function, we can rewrite it using a Pythagorean Identity. In this case, we have two terms containing squares of trigonometric functions. We will rewrite \( \cos^2\left( A \right) \) in terms of \( \sin^2\left( A \right) \) so our equation will only have terms involving the sine function.\[ \begin{array}{rrrclcl}
   & & \cos ^2 \left(A\right)-\sin ^2 \left(A\right)-\sin \left(A\right) & = & 0 & \quad & \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & 1 - \sin^2\left( A \right) - \sin ^2 \left(A\right)-\sin \left(A\right) & = & 0 & \quad & \left( \text{Pythagorean Identity} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 1 - 2\sin^2\left( A \right) - \sin\left( A \right) & = & 0 & \quad & \left( \text{combine like terms} \right)\\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 2\sin^2\left( A \right) + \sin\left( A \right) - 1 & = & 0 & \quad & \left( \text{multiply both sides by }-1 \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \sin\left( A \right) \left[ 2\sin\left( A \right) + 1 \right] & = & 0 & \quad & \left( \text{factor out the GCF} \right) \\
   \end{array} \nonumber \]Thus, by the Zero Factor Property,\[ \sin\left( A \right) = 0 \quad \text{or} \quad \sin\left( A \right) = -\dfrac{1}{2}. \nonumber \]

   For \( \sin\left( A \right) = 0 \)

   From part (a), we know the solutions on \( \left[ 0,2\pi \right) \) are \( x = 0 \) and \( x = \pi \).

   For \( \sin\left( A \right) = -\frac{1}{2} \)

   The sine is negative in \( \mathrm{QIII} \) and \( \mathrm{QIV} \), and the reference angle is \( \frac{\pi}{6} \). Therefore, the solutions on \( \left[ 0,2\pi \right) \) are \( x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \).
   
   Thus, the solution set to this equation on the interval \( \left[ 0,2\pi \right) \) is\[ x \in \left\{ 0, \pi, \dfrac{7\pi}{6}, \dfrac{11\pi}{6} \right\}.\nonumber \]

Advice: Squares on Trigonometric Functions Are Clues

In Examples \( \PageIndex{ 1b } \) and \( \PageIndex{ 1c } \), we used a Pythagorean Identity to transform the square of a trigonometric function into a different trigonometric function. This should convince you that squares on trigonometric functions are often desirable because they allow us to rewrite an expression in an equivalent form using a different trigonometric function.

Checkpoint \( \PageIndex{ 1 } \)

Solve \(1-\sin ^2 \theta+\cos \theta=0\) for \(0^{\circ} \leq \theta \lt 360^{\circ}\).

Answer

\(90^{\circ}, 180^{\circ}, 270^{\circ}\)

Solving Equations Involving Trigonometric Functions Having Different Angular Frequencies

Suppose an equation contains more than one trigonometric function (whether the same or not) but with different angular frequencies. In that case, we will need to use the Double-Angle Identities, Half-Angle Identities, or, in rare cases, the Sum-to-Product or Product-to-Sum Identities to help simplify the equation into the form of Equation \( \ref{trigeqn} \).

Example \(\PageIndex{2}\)

Solve the following equations and list the solutions in the interval \([0,2\pi)\).

1. \(\cos(2x) = 3\cos(x) - 2\)
2. \(\cos(3x) = \cos(5x)\)
3. \(\sin(2x) =\sqrt{3} \cos(x)\)
4. \(\sin(x)\cos\left(\frac{x}{2}\right) + \cos(x)\sin\left(\frac{x}{2}\right) = 1\)
5. \( \sin\left( \frac{x}{2} \right) - \cos\left( x \right) = 0 \)

Solutions

1. In the equation \(\cos(2x) = 3\cos(x) - 2\), we have the same circular function, namely cosine, on both sides but the arguments differ. Using the Double-Angle Identity\[\cos(2x) = 2\cos^{2}(x) - 1,\nonumber\]we obtain a "quadratic in disguise" and proceed as we have done in the past.\[\begin{array}{rrrclcl}
   & & \cos(2x) & = & 3\cos(x) - 2 & & \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & 2\cos^{2}(x) -1 & = & 3\cos(x) -2 & \quad & \left( \text{Double-Angle Identity} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 2\cos^{2}(x) - 3\cos(x) +1 & = & 0 & \quad & \left( \text{gathering terms on the left side} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 2 u^2 - 3 u + 1 & = & 0 & \quad & \left( u-\text{Substitution} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & (2u - 1)(u - 1) & = & 0 & \quad & \left( \text{factor} \right)\\
   \end{array}\nonumber\]This gives \(u = \frac{1}{2}\) or \(u = 1\). Since \(u = \cos(x)\), we get \(\cos(x) = \frac{1}{2}\) or \(\cos(x) = 1\).

   For \( \cos(x) = \frac{1}{2} \)

   Cosine is positive in \( \mathrm{QI} \) and \( \mathrm{QIV} \) and the reference angle is \(\hat{x} = \frac{\pi}{3}\). Therefore, we have the following:\[ \begin{array}{rccl}
   \text{Solutions associated with QI:} & x & = & \frac{\pi}{3} + 2 \pi k \\
   \text{Solutions associated with QIV:} & x & = & \frac{5 \pi}{3} + 2 \pi k, \\
   \end{array} \nonumber \]where \( k \in \mathbb{Z} \).

   For \( \cos\left( x \right) = 1 \)

   This implies \( x \) is a quadrantal angle. Quickly sketching the cosine curve,

   

   we see that \( \cos(x) = 1 \) when \( x = 2 \pi k \).
   
   Thus, the answers which lie in \([0,2\pi)\) are \(x =0\), \(x = \frac{\pi}{3}\), and \(x = \frac{5\pi}{3}\).
2. The given equation, \(\cos(3x) = \cos(5x)\), is troublesome for a few reasons. First, while the equation involves a single trigonometric function (which happens to occur twice), the angular frequency of each function is different. The second issue is that, if we use our previous tactic of moving all trigonometric functions to one side, we get\[ \cos\left( 3x \right) - \cos\left( 5x \right) = 0, \nonumber \]and there is nothing from Algebra that can help us (e.g., there isn't a common factor). Finally, we cannot easily use the Double-Angle Identities for the cosine because the angular frequency of each function is not even.² What do we do?
   
   If you have exhausted all hope of rewriting your equation using algebraic manipulations and fundamental trigonometric identities, then you likely need to reach into your "abstract bag of tricks." Recall that I told you not to memorize the Sum-to-Product and Product-to-Sum Identities; however, I did state that you needed to know they exist. This is because, every once in a while, you will run into a need for them. Specifically, in this case, we need the following Sum-to-Product Identity:\[ \cos\left( \alpha \right) - \cos\left( \beta \right) = -2 \sin\left( \dfrac{\alpha + \beta}{2} \right) \sin\left( \dfrac{\alpha - \beta}{2} \right). \nonumber \]Therefore, our solution process goes like this:\[ \begin{array}{rrrclcl}
   & & \cos\left( 3x \right) & = & \cos\left( 5x \right) & & \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \cos\left( 3x \right) - \cos\left( 5x \right) & = & 0 & \quad & \left( \text{variables expressions to one side} \right) \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & - 2 \sin\left( \dfrac{3x + 5x}{2}\right)\sin\left( \dfrac{3x - 5x}{2}\right) & = & 0 & \quad & \left( \text{Sum-To-Product Identities} \right) \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & - 2 \sin\left( 4x\right)\sin\left( -x\right) & = & 0 & \quad & \left( \text{simplify} \right) \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & 2 \sin\left( 4x\right)\sin\left( x\right) & = & 0 & \quad & \left( \text{Symmetry Identities} \right) \\
   \end{array} \nonumber \]Hence, the equation \(\cos(3x) = \cos(5x)\) is equivalent to \(2 \sin(4x) \sin(x) = 0\). From this, we get \(\sin(4x) = 0\) or \(\sin(x)\) = 0.

   For \( \sin(4x) = 0 \)

   From the base graph for the sine,

   

   we know the sine is zero when its argument is an integer multiple of \( \pi \). That is,\[ 4x = \pi k, \nonumber \]where \( k \in \mathbb{Z} \). This means\[ x = \dfrac{\pi}{4}k. \nonumber \]On the interval \( \left[ 0,2\pi \right) \), this leads to the solutions \(x = 0\), \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), \(\frac{3\pi}{4}\), \(\pi\), \(\frac{5\pi}{4}\), \(\frac{3\pi}{2}\) and \(\frac{7\pi}{4}\).

   For \( \sin(x) = 0 \)

   We know the solutions on \( \left[ 0,2\pi \right) \) are \( x=0 \) and \( \pi \); however, these are already represented in the list of solutions for \(\sin(4x) = 0\).
   
   Thus, the solutions to the original equation on the interval \([0,2\pi)\) are\[x = 0, \, \dfrac{\pi}{4}, \, \dfrac{\pi}{2}, \, \dfrac{3\pi}{4}, \, \pi, \, \dfrac{5\pi}{4}, \, \dfrac{3\pi}{2}, \, \text{ and } \, \dfrac{7\pi}{4}. \nonumber \]
3. Here is where I shorten the discussion and rely on your ability to follow the logic we have been using up to this point.\[\begin{array}{rrrclcl}
   & & \sin\left( 2x \right) & = & \sqrt{3} \cos\left( x \right) & & \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \sin(2x) - \sqrt{3} \cos(x) & = & 0 & \quad & \left( \text{move variable terms to one side} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & 2 \sin(x) \cos(x) - \sqrt{3} \cos(x) & = & 0 & \quad & \left( \text{Double-Angle Identity} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \cos(x) \left(2 \sin(x) - \sqrt{3}\right) & = & 0 & \quad & \left( \text{factor out the GCF} \right) \\
   \end{array}\nonumber\]From this, we get \(\cos(x) = 0\) or \(\sin(x) = \frac{\sqrt{3}}{2}\).

   For \(\cos(x) = 0\)

   Since \( 0 \) is a quadrantal ratio for the cosine, we quickly sketch the cosine function.

   

   We see the solutions are \(x = \frac{\pi}{2} + \pi k\) for \(k \in \mathbb{Z}\). On this interval \( \left[ 0,2\pi \right) \), this gives the restricted solutions \( x=\frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

   For \(\sin(x) = \frac{\sqrt{3}}{2}\)

   The sine is positive in \( \mathrm{QI} \) and \( \mathrm{QII} \), and the reference angle for sine with the ratio \( \frac{\sqrt{3}}{2} \) is \( \hat{x} = \frac{\pi}{3} \). Thus, we get the solutions \(x = \frac{\pi}{3} + 2\pi k\) or \(x = \frac{2\pi}{3} + 2\pi k\) for \(k \in \mathbb{Z}\).³
   
   The answers which lie in \([0,2\pi)\) are \(x = \frac{\pi}{2}\), \(\frac{3\pi}{2}\), \(\frac{\pi}{3}\) and \(\frac{2\pi}{3}\).
4. There seems to be no quick way to get the trigonometric functions or their arguments to match in the equation \(\sin(x)\cos\left(\frac{x}{2}\right) + \cos(x)\sin\left(\frac{x}{2}\right) = 1\); however, if we stare at it long enough, we realize that the left side is the expanded form of the Sum of Angles Identity for the sine.\[ \begin{array}{rrrclcl}
   & & \sin(x)\cos\left(\dfrac{x}{2}\right) + \cos(x)\sin\left(\dfrac{x}{2}\right) & = & 1 & & \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & \sin\left(x + \dfrac{x}{2}\right) & = & 1 & \quad & \left( \text{Sum of Angles Identity} \right) \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \sin\left( \dfrac{3}{2}x \right) & = & 1 & \quad & \left( \text{combine like terms} \right) \\
   \end{array} \nonumber \]Since \( 1 \) is a quadrantal ratio for the sine, we can quickly sketch \( y = \sin(\theta) \)

   

   to see that \( \sin(\theta) = 1 \) when \( \theta = \frac{\pi}{2} + 2\pi k \), where \( k \in \mathbb{Z} \). Therefore,\[ \dfrac{3}{2}x = \dfrac{\pi}{2} + 2 \pi k \implies x = \dfrac{\pi}{3} + \dfrac{4\pi}{3} k. \nonumber \]Two of these solutions lie in \([0,2\pi)\): \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\).
5. As in part (c), we have two different trigonometric functions with different angular frequencies. Moreover, we immediately identify that we are locked from doing any impactful Arithmetic or Algebra; however, in this case, there is a fundamental identity we can use - the Half-Angle Identity for sine.\[ \begin{array}{rrrclcl}
   & & \sin\left( \dfrac{x}{2} \right) - \cos\left( x \right) & = & 0 & & \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & \pm \sqrt{\dfrac{1 - \cos\left( x \right)}{2}} - \cos\left( x \right) & = & 0 & \quad & \left( \text{Half-Angle Identity} \right) \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \pm\sqrt{\dfrac{1 - \cos\left( x \right)}{2}} & = & \cos\left( x \right) & \quad & \left( \text{isolating the radical} \right) \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \dfrac{1 - \cos\left( x \right)}{2} & = & \cos^2\left( x \right) & \quad & \left( \text{squaring both sides } \right. \\
   & & & & & \quad & \left. \implies \textbf{CHECK SOLUTIONS!} \right) \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 1 - \cos\left( x \right) & = & 2\cos^2\left( x \right) & \quad & \left( \text{multiplying both sides by }2 \right) \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 0 & = & 2\cos^2\left( x \right) + \cos\left( x \right) - 1 & \quad & \left( \text{recognizing the equation is quadratic-in-form} \right) \\
   \\
   \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 0 & = & \left( 2\cos\left( x \right) - 1 \right)\left( \cos\left( x \right) + 1 \right) & \quad & \left( \text{factoring} \right)
   \end{array} \nonumber \]Hence, either \( \cos\left( x \right) = \frac{1}{2} \) or \( \cos\left( x \right) = -1 \).

   For \( \cos\left( x \right) = \frac{1}{2} \)

   The cosine function is positive in \( \mathrm{QI} \) and \( \mathrm{QIV} \), and the reference angle is \( \hat{x} = \frac{\pi}{3} \). Therefore, we have the following:\[ \begin{array}{rccl}
   \text{Solutions associated with QI:} & x & = & \frac{\pi}{3} + 2 \pi k \\
   \text{Solutions associated with QIV:} & x & = & \frac{5 \pi}{3} + 2 \pi k, \\
   \end{array} \nonumber \]where \( k \in \mathbb{Z} \). On the interval \( \left[ 0,2\pi \right) \), this equates to \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \); however, since we squared both sides of the equation while solving (see the step with "CHECK SOLUTIONS"), we need to check to see if these solutions work with the original equation.
   
   Letting \( x = \frac{\pi}{3} \) in the original equation, we get\[ \begin{array}{rcl}
   \sin\left( \dfrac{\pi/3}{2} \right) - \cos\left( \dfrac{\pi}{3} \right) & = & \sin\left( \dfrac{\pi}{6} \right) - \cos\left( \dfrac{\pi}{3} \right) \\
   \\
   & = & \dfrac{1}{2} - \dfrac{1}{2} \\
   \\
   & = & 0 \\
   \\
   & = & \textbf{RHS} \\
   \end{array} \nonumber \]Thus, \( x = \frac{\pi}{3} \) is a solution.
   
   Trying \( x = \frac{5\pi}{3} \), we get\[ \begin{array}{rcl}
   \sin\left( \dfrac{5\pi/3}{2} \right) - \cos\left( \dfrac{5\pi}{3} \right) & = & \sin\left( \dfrac{5\pi}{6} \right) - \cos\left( \dfrac{5\pi}{3} \right) \\
   \\
   & = & \dfrac{1}{2} - \dfrac{1}{2} \\
   \\
   & = & 0 \\
   \\
   & = & \textbf{RHS} \\
   \end{array} \nonumber \]Thus, \( x = \frac{5\pi}{3} \) is also a solution.

   For \( \cos\left( x \right) = -1 \)

   From the graph of the cosine,

   

   we see that the solutions to this equation occur at \( x = \pi + 2\pi k \), where \( k \in \mathbb{Z} \). On the interval \( \left[ 0,2\pi \right) \), this gives \( x = \pi \). Again, we must check this solution.
   
   Letting \( x = \pi \) in the original equation, we get\[ \begin{array}{rcl}
   \sin\left( \dfrac{\pi}{2} \right) - \cos\left( \pi \right) & = & 1 - (-1) \\
   \\
   & = & 2 \\
   \\
   & \neq & \textbf{RHS} \\
   \end{array} \nonumber \]Thus, \( x = \pi \) is an extraneous solution, and we throw it out.
   
   In the end, our solutions on the interval \(\left[ 0,2\pi \right)\) to the original equation are \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \).

Instructors will tell you to "check your solutions" after solving an equation. While this is always great advice, let's be realistic - most people don't check their solutions. With that sense of realism in mind, I want to clarify something from Algebra and tell you when you must check the validity of your solutions for the risk of them being wrong. There are three general cases when you must check solutions after solving an equation:

1. when the equation contains a function with a restricted domain,
2. when you multiply both sides of an equation by a variable expression, and
3. when you apply an even power (e.g., a square) to both sides of an equation.

When solving trigonometric equations, you can run into any of these issues. In Example \( \PageIndex{ 2e } \), we ended up squaring both sides of the equation in the process of finding the solution (see the step with "CHECK SOLUTIONS!"). When doing so, you risk introducing "false" solutions - called extraneous solutions. Indeed, that is precisely what happened!

Checkpoint \(\PageIndex{2}\)

Find all solutions to the equation on the interval \( \left[ 0,2\pi \right) \).\[ \sec^2\left( x \right) - 6 = 4 \tan\left( x \right) \nonumber \]

Answer

\( x = \tan^{-1}(5) \), \( x = \frac{3\pi}{4} \), \( x = \pi + \tan^{-1}(5) \), and \( x = \frac{7\pi}{4} \)

Solving Other Trigonometric Equations

At any given step while solving an equation, the tactic I have continually modeled throughout this material is to do the Arithmetic first (i.e., performing simple operations on constants). Once this is done, you move on to see if there is any Algebra to be done (e.g., combining like terms, factoring, etc.). Once you have exhausted all possibilities to use Arithmetic and Algebra, then (and only then) you move on to looking for identities or concepts from Trigonometry.

But what happens when you exhaust all three and you're stuck?

In truth, this means one of two things - either the equation you were given involves a combination of trigonometric functions and non-trigonometric functions (in which case you have no choice but to use technology), or you missed some not-so-obvious Algebra or Trigonometry (Example \( \PageIndex{ 2b } \), which required the abstract Sum-to-Product Identities, is an excellent example of this). There is no way to cover all possible "oddball" cases. Still, the following example covers a case that comes up often enough that we should investigate it.

Example \(\PageIndex{3}\)

Solve the equation and list the solutions that lie in the interval \([0,2\pi)\).\[\cos(x) = \sin(x)\nonumber \]

Solutions

The angular frequencies are the same, but the circular functions are different. This has been fine up to this point because all the previous examples involving two different circular functions had at least one of these functions being squared. Squares of trigonometric functions are blessings in disguise, as we can always try to change them using the Pythagorean Identities. In this case, there are no squares. So what do we do? Since we would like squares, why don't we square both sides?⁴\[ \begin{array}{rrrclcl}
& & \cos(x) & = & \sin(x) & & \\
\\
\scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \cos^2(x) & = & \sin^2(x) & \quad & \left( \text{square both sides} \right. \\
& & & & & & \left. \textbf{CHECK SOLUTIONS!} \right) \\
\scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & 1 - \sin^2(x) & = & \sin^2(x) & \quad & \left( \text{Pythagorean Identity} \right) \\
\\
\scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 1 & = & 2 \sin^2(x) & \quad & \left( \text{move all variable terms to one side} \right) \\
\\
\scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \dfrac{1}{2} & = & \sin^2(x) & \quad & \left( \text{divide both sides by }2 \right) \\
\\
\scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \pm \dfrac{1}{\sqrt{2}} & = & \sin(x) & \quad & \left( \text{Extraction of Roots} \right) \\
\end{array} \nonumber \]Therefore, \( \hat{x} = \frac{\pi}{4} \) and we need to check all quadrants.\[ \begin{array}{rl}
\text{Solutions associated with QI: } & x = \hat{x} + 2\pi k = \dfrac{\pi}{4} + 2 \pi k \\
\\
\text{Solutions associated with QII: } & x = \pi - \hat{x} + 2\pi k = \pi - \dfrac{\pi}{4} + 2 \pi k = \dfrac{3 \pi}{4} + 2 \pi k \\
\\
\text{Solutions associated with QIII: } & x = \pi + \hat{x} + 2\pi k = \pi + \dfrac{\pi}{4} + 2 \pi k = \dfrac{5 \pi}{4} + 2 \pi k \\
\\
\text{Solutions associated with QIV: } & x = 2\pi - \hat{x} + 2\pi k = 2\pi - \dfrac{\pi}{4} + 2 \pi k = \dfrac{7 \pi}{4} + 2 \pi k \\
\end{array} \nonumber \]where \( k \in \mathbb{Z} \). This is a case where, rather than compressing these solutions into two families (which we could do), it's easier to check the four possibilities separately. In all four cases, the reference angles are \( \hat{x} = \frac{\pi}{4} \). Therefore, the sine and cosine functions become \( \pm \frac{1}{\sqrt{2}} \). We just require the signs to be the same.

In \(\mathrm{QI}\), \( \cos{\left( \frac{\pi}{4} \right)} = \frac{1}{\sqrt{2}} = \sin{\left( \frac{\pi}{4} \right)} \). In \(\mathrm{QIII}\), \( \cos{\left( \frac{5\pi}{4} \right)} = -\frac{1}{\sqrt{2}} = \sin{\left( \frac{5\pi}{4} \right)} \); however, we can throw out the other two possibilities because the signs of the functions are opposite in both \(\mathrm{QII}\) and \(\mathrm{QIV}\). Thus, the solution set on the interval \( \left[ 0, 2\pi \right) \) is \( x = \frac{\pi}{4}\) and \(x = \frac{5 \pi}{4} \).

Another option with Example \( \PageIndex{ 3 } \) is to start by dividing both sides by \( \cos\left( x \right) \). Doing so, we get\[ 1 = \tan\left( x \right). \nonumber \]This is much easier to solve, however, by dividing by the cosine, we have introduced a possible issue - what if \( \cos\left( x \right) = 0 \)?

A little thought goes a long way here. If \( \cos\left( x \right) = 0 \), then \( x = \frac{\pi}{2} + \pi k \). If we go back to the original equation and plug in \( \frac{\pi}{2} \), we would get\[ \begin{array}{rrcl}
& \sin\left( \dfrac{\pi}{2} \right) & = & \cos\left( \dfrac{\pi}{2} \right) \\
\implies & 1 & = & 0 \\
\end{array} \nonumber \]Since this is obviously not true, \( x = \frac{\pi}{2} \) is not a solution to the equation. It is a simple task to verify that \( x = \frac{3\pi}{2} \) is also not a solution. Therefore, we don't have to worry about \( \cos\left( x \right) = 0 \), and we can go back to solving\[ \tan\left( x \right) = 1. \nonumber \]Tangent is positive in \( \mathrm{QI} \) and \( \mathrm{QIII} \), and \( \hat{x} = \frac{\pi}{4} \). Hence, the solutions are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

Combining Waves Having Equal Angular Frequencies

Recall that a sinusoidal function of the form \(A\sin (Bx+\phi)\) can be rewritten using the Sum of Angles Identity. That is,\[\begin{array}{rclcl}
A\,\sin \left(Bx+\phi\right) & = & A\left[\sin (Bx)\cos (\phi)+\cos (Bx)\sin (\phi)\right] & \quad & \left( \text{Sum of Angles Identity} \right) \\
& = & A\left[\cos (\phi)\sin (Bx)+\sin (\phi)\cos (Bx)\right] & \quad & \left( \text{rearrange terms} \right) \\
& = & A \,\cos(\phi) \sin(Bx) + A \,\sin(\phi) \cos(Bx) & \quad & \left( \text{distribute} \right) \\
& = & m \, \sin(Bx) + n \, \cos(Bx) & \quad & \left( \text{substitute }m = A \, \cos\left( \phi \right) \text{ and } n = A\, \sin\left( \phi \right) \right) \\
\end{array} \label{sumoftrig}\]Notice that the result of \ref{sumoftrig}, \( m \, \sin(Bx) + n \, \cos(Bx) \), is a vertical stretch (or compression) of the sine added to a different vertical stretch (or compression) of the cosine, but both have the same angular frequency.

Looking at the work in \ref{sumoftrig}, but in reverse order, we reach a fascinating conclusion - if you have a function that is the sum (or difference) of a sine and cosine function having the same angular frequencies, the result is a single sine function with the same angular frequency, but having a phase shift. This is true even if the original sine and cosine functions have different amplitudes.

The question that we should have is if given a function of the form\[f(x) = m\,\sin (Bx)+n\,\cos (Bx),\nonumber \]how do we find \( A \) and \( \phi \) to write this as\[ f(x) = A\,\sin\left( Bx + \phi \right)? \nonumber \]The clue to answering this question is in the last line of \ref{sumoftrig}. In that step, we let \( m = A \, \cos\left( Bx \right) \) and \( n = A \, \sin\left( Bx \right) \). Therefore, we can find the value of \( A \) by using the following ingenious move.\[ \begin{array}{rclcl}
m^2 + n^2 & = & A^2 \cos^2\left( Bx \right) + n^2 \sin^2\left( Bx \right) & \quad & \left( \text{substitute} \right) \\
& = & A^2 \left[ \cos^2\left( Bx \right) + \sin^2\left( Bx \right) \right] & \quad & \left( \text{factor} \right) \\
& = & A^2 \cdot [1] & \quad & \left( \text{Pythagorean Identity} \right) \\
& = & A^2 & & \\
\end{array}\nonumber \]Thus, we will choose\[ A = \sqrt{m^2 + n^2}. \nonumber \]Note that we will deliberately forgo the possibility of \( A \) being negative.

Now that we have a value for \( A \), we still need to find the value of \( \phi \) so that we can properly convert \( f(x) = m\,\sin (Bx)+n\,\cos (Bx) \) into \( f(x) = A\,\sin\left( Bx + \phi \right) \). Again, the clue to finding \( \phi \) is in \ref{sumoftrig}. In the last line, we made the substitutions\[ m = A \, \cos\left( \phi \right) \text{ and } n = A\, \sin\left( \phi \right). \nonumber \]This means\[ \cos\left( \phi \right) = \dfrac{m}{A} \text{ and } \sin\left( \phi \right) = \dfrac{n}{A}. \nonumber \]Since \( A = \sqrt{m^2 + n^2} \) is positive, we can determine the quadrant in which \( \phi \) terminates using the signs of \( m \) and \( n \) and how those affect the quadrant choices for \( \cos\left( \phi \right) \) and \( \sin\left( \phi \right) \), respectively. Once we have the proper quadrant for \( \phi \), we find the reference angle, \( \widehat{\phi} \), using an inverse trigonometric function (arcsine, arccosine, or arctangent all work fine). 

The following theorem summarizes these results.

Theorem

\[ m \, \sin\left( Bx \right) + n \, \cos\left( Bx \right) = A \, \sin\left( Bx + \phi \right), \nonumber \]where\[ A = \sqrt{m^2 + n^2} \nonumber \]and\[ \cos\left( \phi \right) = \dfrac{m}{A} \text{ and } \sin\left( \phi \right) = \dfrac{n}{A}. \nonumber \]

It's time we showcase the method using a concrete example.

Example \(\PageIndex{4}\)

Rewrite \(4\sqrt{3} \sin (2x)-4\cos (2x)\) as a single sinusoidal function.

Solution

Let \( m = 4\sqrt{3} \) and \( n = -4 \). Therefore,\[A =\sqrt{\left(4\sqrt{3} \right)^{2} +\left(-4\right)^{2}} = \sqrt{16\cdot 3+16} = \sqrt{64} = 8.\nonumber \]We now turn to finding \( \phi \).We know\[ \cos\left( \phi \right) = \dfrac{m}{A} = \dfrac{4\sqrt{3}}{8} = \dfrac{\sqrt{3}}{2} \quad \text{and} \quad \sin\left( \phi \right) = \dfrac{n}{A} = \dfrac{-4}{8} = -\dfrac{1}{2}. \nonumber \]Cosine is positive in \( \mathrm{QI} \) and \( \mathrm{QIV} \), and sine is negative in \( \mathrm{QIII} \) and \( \mathrm{QIV} \). The intersection of these choices is \( \mathrm{QIV} \). Thus, \( \phi \in \mathrm{QIV} \).

We can use either relation to find the reference angle, but I will use the cosine. The reference angle for cosine that returns \( \frac{\sqrt{3}}{2} \) is \( \widehat{\phi} = \frac{\pi}{6} \). Therefore, \( \phi = \frac{11\pi}{6} \).

Combining these results gives us the expression\[8\sin \left(2x+\dfrac{11\pi }{6} \right).\nonumber\]

Checkpoint \(\PageIndex{4}\)

Rewrite \(-3\sqrt{2} \sin (5x)+3\sqrt{2} \cos (5x)\) as a single sinusoidal function.

Answer

\(6\sin \left(5x+\dfrac{3\pi }{4} \right)\)

Rewriting a combination of sine and cosine of equal periods as a single sinusoidal function provides an alternative approach for solving some equations.

Example \(\PageIndex{5}\)

Find all solutions to the equation \(3\sin (2x)+4\cos (2x)=1\) on the interval \( \left[ 0,2\pi \right) \).

Solution

There are two ways to approach this problem: recognizing that you could isolate one of the trigonometric functions and square both sides or using what we just learned (combining sine and cosine into a single sinusoidal function). I encourage you to try the first method on your own (it slightly mimics Example \( \PageIndex{ 3 } \), but it is more challenging). For now, we will pretend that we cannot see the possibility of using Algebra in the initial equation.

Not having any Arithmetic or Algebra to do, we then recall what we just learned. We let\[ A = \sqrt{\left( 3 \right)^2 + \left( 4 \right)^2} = \sqrt{25} = 5,  \nonumber \]where\[ \cos\left( \phi \right) = \dfrac{3}{5} \quad \text{and} \quad \sin\left( \phi \right) = \dfrac{4}{5}. \nonumber \]The sine and cosine are both positive in \( \mathrm{QI} \). Therefore, \( \phi \in \mathrm{QI} \). Finally,\[ \widehat{\phi} = \cos^{-1}\left( \left| \dfrac{3}{5} \right| \right) = \cos^{-1}\left( \dfrac{3}{5}  \right). \nonumber \]Hence,\[ \phi = \cos^{-1}\left( \dfrac{3}{5} \right). \nonumber \]We now can rewrite the left side of the original equation as \( A \, \sin\left( 2x + \phi \right) \) and get\[ 5 \sin\left( 2x + \cos^{-1}\left( \dfrac{3}{5} \right) \right) = 1. \nonumber \]This is a much easier equation to solve.\[ \begin{array}{rrrclcl}
&  &  5 \sin\left( 2x + \cos^{-1}\left( \dfrac{3}{5} \right) \right) & = & 1 & & \\
\\
\scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \sin\left( 2x + \cos^{-1}\left( \dfrac{3}{5} \right) \right) & = & \dfrac{1}{5} & \quad & \left( \text{divide both sides by }5 \right) \\
\end{array} \nonumber \]Sine is positive in \( \mathrm{QI} \) and \( \mathrm{QII} \) and the reference angle is \( \widehat{\theta} = \sin^{-1}\left( \frac{1}{5} \right) \), where \( \theta = 2x + \cos^{-1}\left( \frac{3}{5} \right) \). Therefore, we get:\[ \begin{array}{rl}
\text{Solutions associated with QI: } & 2x + \cos^{-1}\left( \dfrac{3}{5} \right) = \sin^{-1}\left( \dfrac{1}{5} \right) + 2 \pi k \\
\\
\text{Solutions associated with QII: } & 2x + \cos^{-1}\left( \dfrac{3}{5} \right) = \pi - \sin^{-1}\left( \dfrac{1}{5} \right) + 2 \pi k \\
\end{array} \nonumber \]for \( k \in \mathbb{Z} \). Solving each equation for \( x \), we get:\[ \begin{array}{rl}
\text{Solutions associated with QI: } & x = \dfrac{1}{2}\sin^{-1}\left( \dfrac{1}{5} \right) + \pi k - \dfrac{1}{2}\cos^{-1}\left( \dfrac{3}{5} \right) \approx -0.362968649 + \pi k \\
\\
\text{Solutions associated with QII: } & 2x + \cos^{-1}\left( \dfrac{3}{5} \right) = \dfrac{\pi}{2} - \dfrac{1}{2}\sin^{-1}\left( \dfrac{1}{5} \right) + \pi k - \dfrac{1}{2}\cos^{-1}\left( \dfrac{3}{5} \right) \approx 1.006469757 + \pi k \\
\end{array} \nonumber \]We only want solutions on \( \left[ 0,2\pi \right) \), so we build a quick table.\[ \begin{array}{|c|c|c|}
\hline \mathbf{k} & \mathbf{-0.362968649 + \pi k} & \mathbf{1.006469757 + \pi k} \\
\hline 0 & -0.362968649 & 1.006469757 \\
\hline 1 & 2.778624005 & 4.148062411 \\
\hline 2 & 5.920216658 & 7.289655065 \\
\hline 3 & 9.061809312 & \vdots \\
\hline \end{array} \nonumber \]Of these results, the solutions on the requested interval are \( x \approx 1.0065 \), \( x \approx 2.7786 \), \( x \approx 4.1481 \), and \( x \approx 5.9202 \).

Checkpoint \(\PageIndex{5}\)

Find all solutions of the equation on the interval \( \left[ 0,2\pi \right) \).\[ \cos\left( 2t \right) - \sqrt{3} \sin\left( 2t \right) = 1 \nonumber \]

Answer

\( x = 0 \), \( x = \frac{2\pi}{3} \), \( x = \pi \), and \( x = \frac{5\pi}{3} \).

---

Footnotes

¹  Recall from Algebra that when you divide both sides of an equation by a variable expression, you are assuming the expression you are dividing by is nonzero. Therefore, in the given equation, unless you can guarantee that \( \sin^2\left( x \right) \neq 0 \), you cannot divide both sides of the equation by \( \sin^2\left( x \right) \).

² Technically, we can use the Double-Angle Identities, however, they would not be very helpful. This is because, for example,\[ \cos\left( 3x \right) = \cos\left( 2\left( \dfrac{3}{2}x \right) \right) = 2\cos^2\left( \dfrac{3}{2}x \right) - 1, \nonumber \]and this looks much worse than the original expression.

³ Notice that I am speeding up descriptions as we move through. That is, the amount of detail I describe once a topic is introduced becomes less and less. This is purposeful. You need to be fluent in this material so it becomes second nature.

⁴  Remember, when you square both sides of an equation, you run the risk of introducing extraneous solutions. Therefore, you must always check your solutions when you perform this action.

---

Homework

Concept Check

1. Will there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not.

2. When solving a trigonometric equation involving more than one trigonometric function, do we always want to try to rewrite it so it is expressed in one trigonometric function? Why or why not?

Basic Skills

For Problems 3 - 57, solve the equation, giving exact solutions on the interval \( \left[ 0,2\pi \right) \).

3. \(\sin\left( 2\theta \right)+\sqrt{2} \cos\left( \theta \right)=0\)

4. \(\sin\left( 2\alpha \right) \sin\left( \alpha \right)=\cos\left( \alpha \right)\)

5. \(\cos\left( 2t \right)-5 \cos\left( t \right)+3=0\)

6. \(\cos\left( 2x \right)+3 \sin\left( x \right)=2\)

7. \(\tan\left( 2\beta \right)+2 \sin\left( \beta \right)=0\)

8. \(\tan\left( 2z \right)-2 \cos\left( z \right)=0\)

9. \(3 \cos\left( \phi \right)-\sin \left(\frac{\pi}{2}-\phi\right)=\sqrt{3}\)

10. \(\sin\left( w \right)+\cos \left(\frac{\pi}{2}-w\right)=1\)

11. \(\sin\left( 2\phi \right) \cos\left( \phi \right)+\cos\left( 2\phi \right) \sin\left( \phi \right)=1\)

12. \(\cos\left( \theta \right) \cos\left( 3\theta \right)+\sin\left( \theta \right) \sin\left( 3\theta \right)=\frac{\sqrt{2}}{2}\)

13. \(\sin \left( x \right) = \cos \left( x \right)\)

14. \(\sin \left( 2x \right) = \sin \left( x \right)\)

15. \(\sin \left( 2x \right) = \cos \left( x \right)\)

16. \(2\tan^2\left( t \right)=3\sec\left( t \right)\)

17. \(\cos \left( 2x \right) = \sin \left( x \right)\)

18. \(\cos \left( 2x \right) = \cos \left( x \right)\)

19. \(\cos\left( 2x \right) = 2 - 5\cos\left( x \right)\)

20. \(3\cos\left( 2x \right) + \cos\left( x \right) + 2 = 0\)

21. \(\cos\left( 2x \right) = 5\sin\left( x \right) - 2\)

22. \(3\cos\left( 2x \right) = \sin\left( x \right) + 2\)

23. \(2\sec^2\left( x \right) = 3 - \tan\left( x \right)\)

24. \(\tan^2\left( x \right) = 1-\sec\left( x \right)\)

25. \(\cot^2\left( x \right) = 3\csc\left( x \right) - 3\)

26. \(\sec\left( x \right) = 2\csc\left( x \right)\)

27. \(\cos\left( x \right)\csc\left( x \right)\cot\left( x \right) = 6-\cot^2\left( x \right)\)

28. \(\sin\left( 2x \right) = \tan\left( x \right)\)

29. \(\cot^4\left( x \right) = 4\csc^2\left( x \right) - 7\)

30. \(\cos\left( 2x \right) + \csc^2\left( x \right) = 0\)

31. \(\tan^{3} \left( x \right) = 3\tan \left( x \right)\)

32. \(\cos\left( 6x \right)-\cos\left( 3x \right)=0\)

33. \(\sin^2\left( x \right)-\cos^2\left( x \right)-\sin\left( x \right)=0\)

34. \(\tan^{2} \left( x \right) = \frac{3}{2} \sec \left( x \right)\)

35. \(2\sin\left( x \right)\cos\left( x \right)-\sin\left( x \right)+2\cos\left( x \right)-1=0\)

36. \(\cos^{3} \left( x \right) = -\cos \left( x \right)\)

37. \(\tan\left( 2x \right) - 2\cos\left( x \right) = 0\)

38. \(\csc^3\left( x \right) + \csc^2\left( x \right) = 4\csc\left( x \right) + 4\)

39. \(2\tan\left( x \right) = 1 - \tan^2\left( x \right)\)

40. \(\tan \left( x \right) = \sec \left( x \right)\)

41. \(\sin\left( 6x \right) \cos\left( x \right) = -\cos\left( 6x \right) \sin\left( x \right)\)

42. \(\sin\left( 3x \right)\cos\left( x \right) = \cos\left( 3x \right) \sin\left( x \right)\)

43. \(\cos\left( 2x \right)\cos\left( x \right) + \sin\left( 2x \right)\sin\left( x \right) = 1\)

44. \(\cos\left( 5x \right)\cos\left( 3x \right) - \sin\left( 5x \right)\sin\left( 3x \right) = \frac{\sqrt{3}}{2}\)

45. \(\sin\left( x \right) + \cos\left( x \right) = 1\)

46. \(\sin\left( x \right) + \sqrt{3} \cos\left( x \right) = 1\)

47. \(\sqrt{2} \cos\left( x \right) - \sqrt{2} \sin\left( x \right) = 1\)

48. \(\sqrt{3} \sin\left( 2x \right) + \cos\left( 2x \right) = 1\)

49. \(\cos\left( 2x \right) - \sqrt{3} \sin\left( 2x \right) = \sqrt{2}\)

50. \(9\cos\left( 2\theta \right)=9\cos^2\left( \theta \right) -4\)

51. \(3\sqrt{3}\sin\left( 3x \right) - 3\cos\left( 3x \right) = 3\sqrt{3}\)

52. \(\cos\left( 3x \right) = \cos\left( 5x \right)\)

53. \(\cos\left( 4x \right) = \cos\left( 2x \right)\)

54. \(\sin\left( 5x \right) = \sin\left( 3x \right)\)

55. \(\cos\left( 5x \right) = -\cos\left( 2x \right)\)

56. \(\sin\left( 6x \right) + \sin\left( x \right) = 0\)

57. \(\tan\left( x \right) = \cos\left( x \right)\)

For Problems 58 - 65, solve the equation for \(0^{\circ} \leq \theta \leq 360^{\circ}\). Round angles to three decimal places if necessary.

58. \(\cos\left( \theta \right)-\sin^2\left( \theta \right)+1=0\)

59. \(4 \sin\left( \theta \right)+2 \cos^2\left( \theta \right)-3=-1\)

60. \(1-\sin\left( \theta \right)-2 \cos^2\left( \theta \right)=0\)

61. \(3 \cos^2\left( \theta \right)-\sin^2\left( \theta \right)=2\)

62. \(2 \cos\left( \theta \right) \tan\left( \theta \right)+1=0\)

63. \(\cos\left( \theta \right)-\sin\left( \theta \right)=0\)

64. \(\frac{1}{3} \cos\left( \theta \right)=\sin\left( \theta \right)\)

65. \(5 \sin\left( C \right)=2 \cos\left( C \right)\)

For Problems 66 - 70, find all radian solutions to the equation.

66. \(\sin \left(3x\right)\cos \left(6x\right)-\cos \left(3x\right)\sin \left(6x\right)= -0.9\)

67. \(\sin \left(6x\right)\cos \left(11x\right)-\cos \left(6x\right)\sin \left(11x\right)= -0.1\)

68. \(\cos \left(2x\right)\cos \left(x\right)+\sin \left(2x\right)\sin \left(x\right)=1\)

69. \(\cos \left(5x\right)\cos \left(3x\right)-\sin \left(5x\right)\sin \left(3x\right)=\frac{\sqrt{3} }{2}\)

70. \(6\sin\left( 2t \right)+9\sin\left( t \right)=0\)

For Problems 71 - 74, find all radian solutions to the equation.

71. \(\cos \left(5x\right)=-\cos \left(2x\right)\)

72. \(\sin \left(5x\right)=\sin \left(3x\right)\)

73. \(\cos \left(6\theta \right)-\cos \left(2\theta \right)=\sin \left(4\theta \right)\)

74. \(\cos \left(8\theta \right)-\cos \left(2\theta \right)=\sin \left(5\theta \right)\)

For Problems 75 - 78, rewrite the expression as a single function of the form \(A\,\sin\left(Bx+\phi\right)\).

75. \(4\sin \left(x\right)-6\cos \left(x\right)\)

76. \(-\sin \left(x\right)-5\cos \left(x\right)\)

77. \(5\sin \left(3x\right)+2\cos \left(3x\right)\)

78. \(-3\sin \left(5x\right)+4\cos \left(5x\right)\)

For Problems 79 - 82, find the first two positive (radian) solutions.

79. \(-5\sin \left(x\right)+3\cos \left(x\right)=1\)

80. \(3\sin \left(x\right)+\cos \left(x\right)=2\)

81. \(3\sin \left(2x\right)-5\cos \left(2x\right)=3\)

82. \(-3\sin \left(4x\right)-2\cos \left(4x\right)=1\)

For Problems 83 - 98, solve the equation for \(0^{\circ} \leq x \lt 360^{\circ}\).

83. \( 9 \sin^2\left( \theta \right)-6 \sin\left( \theta \right)=1\)

84. \( 4 \cos^2\left( \theta \right)+4 \cos\left( \theta \right)=1\)

85. \( \sec^2\left( \alpha \right)-2 \sec\left( \alpha \right)-3=0\)

86. \( \csc^2\left( \beta \right)+4 \csc\left( \beta \right)-10=0\)

87. \( \csc^2\left( x \right)+4 \csc\left( x \right)-7=0\)

88. \( 3 \cot^2\left( x \right)-3 \cot\left( x \right)-1=0\)

89. \( 2 \sin^2\left( x \right)=1-\cos\left( x \right)\)

90. \( \cos^2\left( \alpha \right)+4=2 \sin\left( \alpha \right)-3\)

91. \( \cos^2\left( \beta \right)-3 \sin\left( \beta \right)+2 \sin^2\left( \beta \right)=0\)

92. \( \sin^2\left( \theta \right)=2 \cos\left( \theta \right)+3 \cos^2\left( \theta \right)\)

93. \( \sec^2\left( x \right)=2 \tan\left( x \right)+4\)

94. \( 3 \tan^2\left( x \right)=\sec\left( x \right)+2\)

95. \( \cos\left( \alpha \right)+1=2 \cos\left( 2\alpha \right)\)

96. \( \cos\left( 2x \right)-3 \sin\left( x \right)-2=0\)

97. \( \csc^2\left( \theta \right)=\cot\left( \theta \right)+5\)

98. \( \csc\left( \theta \right)+5=2 \cot^2\left( \theta \right)+2\)