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# 6.1: Solving Trigonometric Equations

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An equation involving trigonometric functions is called a trigonometric equation. For example, an equation like

$\tan\;A ~=~ 0.75 ~, \nonumber$

which we encountered in Chapter 1, is a trigonometric equation. In Chapter 1 we were concerned only with finding a single solution (say, between $$0^\circ$$ and $$90^\circ$$). In this section we will be concerned with finding the most general solution to such equations.

To see what that means, take the above equation $$\tan\;A = 0.75$$. Using the $$\fbox{\(\tan^{-1}$$}\) calculator button in degree mode, we get $$A=36.87^\circ$$. However, we know that the tangent function has period $$\pi$$ rad $$= 180^\circ$$, i.e. it repeats every $$180^\circ$$. Thus, there are many other possible answers for the value of $$A$$, namely $$36.87^\circ + 180^\circ$$, $$36.87^\circ - 180^\circ$$, $$36.87^\circ + 360^\circ$$, $$36.87^\circ - 360^\circ$$, $$36.87^\circ + 540^\circ$$, etc. We can write this in a more compact form:

$A ~=~ 36.87^\circ \;+\; 180^\circ k \quad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$

This is the most general solution to the equation. Often the part that says "for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$'' is omitted since it is usually understood that $$k$$ varies over all integers. The general solution in radians would be:

$A ~=~ 0.6435 \;+\; \pi k \quad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$

## Example 6.1

Solve the equation $$\;2\,\sin\;\theta \;+\;1 ~=~ 0$$.

Solution:

Isolating $$\sin\;\theta$$ gives $$\;\sin\;\theta ~=~ -\tfrac{1}{2}$$. Using the $$\fbox{\(\sin^{-1}$$}\) calculator button in degree mode gives us $$\theta = -30^\circ$$, which is in QIV. Recall that the reflection of this angle around the $$y$$-axis into QIII also has the same sine. That is, $$\sin\;210^\circ = -\tfrac{1}{2}$$. Thus, since the sine function has period $$2\pi$$ rad $$= 360^\circ$$, and since $$-30^\circ$$ does not differ from $$210^\circ$$ by an integer multiple of $$360^\circ$$, the general solution is:

$\boxed{\theta ~=~ -30^\circ \;+\; 360^\circ k \quad\text{and}\quad 210^\circ \;+\; 360^\circ k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$

$\boxed{\theta ~=~ -\dfrac{\pi}{6} \;+\; 2\pi k \quad\text{and}\quad \dfrac{7\pi}{6} + 2\pi k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$

For the rest of this section we will write our solutions in radians.

## Example 6.2

Solve the equation $$\;2\cos^2 \;\theta \;-\; 1 ~=~ 0$$.

Solution:

Isolating $$\;\cos^2 \;\theta$$ gives us

$\cos^2 \;\theta ~=~ \frac{1}{2} \quad\Rightarrow\quad \cos\;\theta ~=~ \pm\,\frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta ~=~ \frac{\pi}{4}\;,~\frac{3\pi}{4}\;,~\frac{5\pi}{4}\;,~ \frac{7\pi}{4}~, \nonumber$

and since the period of cosine is $$2\pi$$, we would add $$2\pi k$$ to each of those angles to get the general solution. But notice that the above angles differ by multiples of $$\frac{\pi}{2}$$. So since every multiple of $$2\pi$$ is also a multiple of $$\frac{\pi}{2}$$, we can combine those four separate answers into one:

$\boxed{\theta ~=~ \frac{\pi}{4} \;+\; \frac{\pi}{2}\,k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$

## Example 6.3

Solve the equation $$\;2\,\sec\;\theta ~=~ 1$$.

Solution:

Isolating $$\;\sec\;\theta$$ gives us

$\sec\;\theta ~=~ \frac{1}{2} \quad\Rightarrow\quad \cos\;\theta ~=~ \frac{1}{\sec\;\theta} ~=~ 2~, \nonumber$

which is impossible. Thus, there is $$\fbox{no solution}$$.

## Example 6.4

Solve the equation $$\;\cos\;\theta ~=~ \tan\;\theta$$.

Solution:

The idea here is to use identities to put everything in terms of a single trigonometric function:

\nonumber \begin{align*} \cos\;\theta ~&=~ \tan\;\theta\\ \nonumber \cos\;\theta ~&=~ \frac{\sin\;\theta}{\cos\;\theta}\\ \nonumber \cos^2 \;\theta ~&=~ \sin\;\theta\\ \nonumber 1 \;-\; \sin^2 \;\theta ~&=~ \sin\;\theta\\ \nonumber 0 ~&=~ \sin^2 \;\theta \;+\; \sin\;\theta \;-\; 1 \end{align*} \nonumber

The last equation looks more complicated than the original equation, but notice that it is actually a quadratic equation: making the substitution $$x=\sin\;\theta$$, we have

$x^2 \;+\; x \;-\; 1 ~=~ 0 \quad\Rightarrow\quad x ~=~ \frac{-1 \;\pm\; \sqrt{1 - (4)\,(-1)}}{ 2\,(1)} ~=~ \frac{-1 \;\pm\; \sqrt{5}}{2} ~=~ -1.618\;,~0.618 \nonumber$

by the quadratic formula from elementary algebra. But $$-1.618 < -1$$, so it is impossible that $$\;\sin\theta = x = -1.618$$. Thus, we must have $$\;\sin\;\theta = x = 0.618$$. Hence, there are two possible solutions: $$\theta = 0.666$$ rad in QI and its reflection $$\pi - \theta = 2.475$$ rad around the $$y$$-axis in QII. Adding multiples of $$2\pi$$ to these gives us the general solution:

$\boxed{\theta ~=~ 0.666 \;+\; 2\pi k \quad\text{and}\quad 2.475 \;+\; 2\pi k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$

## Example 6.5

Solve the equation $$\;\sin\;\theta ~=~ \tan\;\theta$$.

Solution:

Trying the same method as in the previous example, we get

\nonumber \begin{align*} \sin\;\theta ~&=~ \tan\;\theta\\ \nonumber \sin\;\theta ~&=~ \frac{\sin\;\theta}{\cos\;\theta}\\ \nonumber \sin\;\theta~\cos\;\theta ~&=~ \sin\;\theta\\ \nonumber \sin\;\theta~\cos\;\theta \;-\; \sin\;\theta ~&=~ 0\\ \nonumber \sin\;\theta~(\cos\;\theta \;-\; 1) ~&=~ 0\\ \nonumber &\Rightarrow\quad \sin\;\theta ~=~ 0 \quad\text{or}\quad \cos\;\theta ~=~ 1\\ \nonumber &\Rightarrow\quad \theta ~=~ 0\;,~\pi \quad\text{or}\quad \theta ~=~ 0\\ \nonumber &\Rightarrow\quad \theta ~=~ 0\;,~\pi~, \end{align*} \nonumber

plus multiples of $$2\pi$$. So since the above angles are multiples of $$\pi$$, and every multiple of $$2\pi$$ is a multiple of $$\pi$$, we can combine the two answers into one for the general solution:

$\boxed{\theta ~=~ \pi k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$

## Example 6.6

Solve the equation $$\;\cos\;3\theta ~=~ \frac{1}{2}$$.

Solution:

The idea here is to solve for $$3\theta$$ first, using the most general solution, and then divide that solution by $$3$$. So since $$\;\cos^{-1} \frac{1}{2} = \frac{\pi}{3}$$, there are two possible solutions for $$3\theta$$: $$3\theta = \frac{\pi}{3}$$ in QI and its reflection $$-3\theta = -\frac{\pi}{3}$$ around the $$x$$-axis in QIV. Adding multiples of $$2\pi$$ to these gives us:

$3\theta ~=~ \pm\,\frac{\pi}{3} \;+\; 2\pi k \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$

So dividing everything by $$3$$ we get the general solution for $$\theta$$:

$\boxed{\theta ~=~ \pm\,\frac{\pi}{9} \;+\; \frac{2\pi}{3} k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$

## Example 6.7

Solve the equation $$\;\sin\;2\theta ~=~ \sin\;\theta$$.

Solution:

Here we use the double-angle formula for sine:

\nonumber \begin{align*} \sin\;2\theta ~&=~ \sin\;\theta\\ \nonumber 2\,\sin\theta~\cos\;\theta ~&=~ \sin\;\theta\\ \nonumber \sin\;\theta~(2\,\cos\;\theta \;-\; 1) ~&=~ 0\\ \nonumber &\Rightarrow\quad \sin\;\theta ~=~ 0 \quad\text{or}\quad \cos\;\theta ~=~ \frac{1}{2}\\ \nonumber &\Rightarrow\quad \theta ~=~ 0\;,~\pi \quad\text{or}\quad \theta ~=~ \pm\,\frac{\pi}{3}\\ \nonumber &\Rightarrow\quad \boxed{\theta ~=~ \pi k \quad\text{and}\quad \pm\,\frac{\pi}{3} \;+\; 2\pi k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \end{align*} \nonumber

## Example 6.8

Solve the equation $$\;2\,\sin\;\theta \;-\; 3\,\cos\;\theta ~=~ 1$$.

Solution

We will use the technique which we discussed in Chapter 5 for finding the amplitude of a combination of sine and cosine functions. Take the coefficients $$2$$ and $$3$$ of $$\;\sin\;\theta$$ and $$\;-\cos\;\theta$$, respectively, in the above equation and make them the legs of a right triangle, as in Figure 6.1.1. Let $$\phi$$ be the angle shown in the right triangle. The leg with length $$3 >0$$ means that the angle $$\phi$$ is above the $$x$$-axis, and the leg with length $$2>0$$ means that $$\phi$$ is to the right of the $$y$$-axis. Hence, $$\phi$$ must be in QI. The hypotenuse has length $$\sqrt{13}$$ by the Pythagorean Theorem, and hence $$\;\cos\;\phi = \frac{2}{\sqrt{13}}$$ and $$\;\sin\;\theta = \frac{3}{\sqrt{13}}$$. We can use this to transform the equation to solve as follows:

\nonumber \begin{align*} 2\,\sin\;\theta \;-\; 3\,\cos\;\theta ~&=~ 1\\ \nonumber \sqrt{13}\,\left( \tfrac{2}{\sqrt{13}}\,\sin\;\theta \;-\; \tfrac{3}{\sqrt{13}}\,\cos\;\theta \right) ~&=~ 1\\ \nonumber \sqrt{13}\,( \cos\;\phi\;\sin\;\theta \;-\; \sin\;\phi\;\cos\;\theta ) ~&=~ 1\\ \nonumber \sqrt{13}\,\sin\;(\theta - \phi) ~&=~ 1\quad\text{(by the sine subtraction formula)}\\ \nonumber \sin\;(\theta - \phi) ~&=~ \tfrac{1}{\sqrt{13}}\\ \nonumber &\Rightarrow\quad \theta - \phi ~=~ 0.281 \quad\text{or}\quad \theta - \phi ~=~ \pi - 0.281 = 2.861\\ \nonumber &\Rightarrow\quad \theta ~=~ \phi \;+\; 0.281 \quad\text{or}\quad \theta ~=~ \phi \;+\; 2.861 \end{align*} \nonumber

Now, since $$\;\cos\;\phi = \frac{2}{\sqrt{13}}$$ and $$\phi$$ is in QI, the most general solution for $$\phi$$ is $$\phi = 0.983 + 2\pi k$$ for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$. So since we needed to add multiples of $$2\pi$$ to the solutions $$0.281$$ and $$2.861$$ anyway, the most general solution for $$\theta$$ is:

\begin{align*} \theta ~&=~ 0.983 \;+\; 0.281 \;+\; 2\pi k\quad\text{and}\quad 0.983 \;+\; 2.861 \;+\; 2\pi k\\ &\Rightarrow\quad \boxed{\theta ~=~ 1.264 \;+\; 2\pi k\quad\text{and}\quad 3.844 \;+\; 2\pi k} \quad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \end{align*} \nonumber

Note: In Example 6.8 if the equation had been $$\;2\,\sin\;\theta \;+\; 3\,\cos\;\theta ~=~ 1$$ then we still would have used a right triangle with legs of lengths $$2$$ and $$3$$, but we would have used the sine addition formula instead of the subtraction formula.

6.1: Solving Trigonometric Equations is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Michael Corral via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.