8.7 Solving Trigonometric Equations
- Page ID
- 158769
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- find all solutions to equations involving trig functions
- use trig identities and other tricks to solve equations involving multiple trig functions
Sometimes students escape high school under the impression that trig can be entirely handled with the trusty \( \sin^{-1}\) button on a calculator, but in calculus courses this is not the case. Any time you try to tackle a trig equation by slapping it with an inverse trig function, you are actually losing a lot of information! This is because, in order to define the inverse functions, we had to pretend most of sine, cosine, and tangent didn't exist. We restricted our domain to a tiny chunk of the \(x\)-axis. In some applications, that may be fine, but in others, all that lost information was important! So now we're going to learn how to find solutions to trig equations as will be needed in future courses. For example, how do you figure out all possible angles \(\theta\) such that "\( \sin \theta - \frac{1}{2} = 0\)" will be a true statement?
To find all solutions to a trigonometric equation,
- Use algebra to manipulate the equation until it consists of one single trig function isolated on one side of the equals.
- Find all the solutions in the usual unit circle (angles between \(0\) and \(2\pi\)).
- List all solutions by adding integer multiples of the period to the solutions.
Give at least four solutions to the equation, and write a formula that will describe all solutions.
- \( \sin \theta - \frac{1}{2} = 0\)
- \( \cos \theta = \sin \theta\)
- \( 2\cos \theta = \cos \theta \)
Solution
1. First, we get the trig term by himself: \( \sin \theta = \frac{1}{2}\). Now we ask ourselves, at what angles does sine turn out to be \( \frac{1}{2}\)? Recalling our unit circle, that will happen if \( \theta \) is \( \frac{\pi}{6}\) or \(\frac{5\pi}{6}\). BUT ALSO, it will happen at any other angle values that are coterminal with those! See four options below:
So \(\frac{\pi}{6}+2\pi = \frac{13\pi}{6}\) is a location where \( \sin \theta = \frac{1}{2}\), and so is \( \frac{\pi}{6}+4\pi = \frac{25\pi}{6} \), and any such angle created by adding a multiple of \(2\pi\) to \( \frac{\pi}{6}\). The same is true for \( \frac{5\pi}{6}\) and his coterminal angles \( \frac{5\pi}{6}+2\pi = \frac{17\pi}{6}\), or \(\frac{5\pi}{6} + 4\pi = \frac{29\pi}{6}\), etc. etc. Also, coterminal negative angles are solutions too! For example, \( -\frac{7\pi}{6}\) is coterminal with \( \frac{5\pi}{6}\) and \( -\frac{11\pi}{6}\) is coterminal with \(\frac{\pi}{6}\). The way to summarize all of these solutions is with formulas like this:
\[ \theta = \frac{\pi}{6} + 2k\pi, \quad \theta = \frac{5\pi}{6}+2k\pi, \quad \text{where \(k\) is any integer} \notag \]
Since \(k\) can be a positive or negative integer, this will cover all our bases.
2. Our first step is to boil everything down to a single trig function isolated on one side of the equation. We might like to divide both sides by \( \cos \theta\) to get \( 1 = \tan \theta\). By doing so, however, we are assuming that \(\cos \theta \neq 0\)! Is that okay? Well, any angle \(\theta\) where \(\cos \theta = 0\), such at \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\), etc. will not be a solution to this equation, because at those angles, \(\sin \theta \neq 0\). So it's okay to make this assumption! Now, when does tangent turn out to be \(1\)? Anywhere on the unit circle where \( \sin \theta\) and \(\cos \theta\) are equal. This happens at \( \frac{\pi}{4}\) and \( \frac{5\pi}{4}\). Our final answer is then \( \theta = \frac{\pi}{4}+2k\pi, \theta = \frac{5\pi}{4} + 2k\pi\), where \(k\) is any integer. You can also just use common sense on this equation instead of doing the isolation step to convert to tangent. Just look at the equation and ask yourself, "Hmm, where are cosine and sine the same?" and ponder your unit circle. Specific solutions we could list out by choosing \(k\) options could be: \(\theta = -\frac{7\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4},\) etc.
3. We bring together the like terms and isolate the trig function to get \( \cos \theta = 0 \). Now, where is cosine zero? At \( \frac{\pi}{2}\) and \(\frac{3\pi}{2}\). Any angle given by \( \theta = \frac{\pi}{2}+2k\pi\) or \( \theta = \frac{3\pi}{2}+2k\pi\) will be a solution. Some specific examples are: \( -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2},\) etc. This can also be described as "odd multiples of \(\frac{\pi}{2}\)," if you like.
Give at least three specific solutions to the equation, and describe all possible solutions.
- \( \sin \theta = 0\)
- \( 2\cos \theta = - \sqrt{2} \)
- \( \tan^2 \theta = 1\)
- Answer
-
- \(\sin \theta\) is zero if \(\theta = -\pi, 0, \pi, 2\pi,\) etc. Aka, \( \theta = 0+2k\pi\) and \(\theta = \pi+2k\pi\), or simplified, any integer multiple of \(\pi\).
- Remember that \( - \frac{\sqrt{2}}{2}\) is another way of writing \( -\frac{1}{\sqrt{2}}\). So \(\theta = \frac{3\pi}{4}+2k\pi\), \(\theta = \frac{5\pi}{4}+2k\pi\), where \(k\) is any integer.
- Hint: if \(\tan \theta = \pm 1\), then the equation \( \tan^2 \theta = 1\) will be true, so you're looking for angles where tangent is \(1\) or \(-1\). So \( \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\) etc. all work! This can be described as all odd multiples of \(\frac{\pi}{4}\). One way you can write this in math code as \( \theta = (2k+1)\frac{\pi}{4}\) (can you see why?) if you want.
Sometimes an equation has multiple trig function terms in it and we need to use identities to manipulate it into an easily solvable form. Here are some more complicated examples. Each one will show you a special technique.
Solve the equations.
- \( 1+ \sin \theta = 1 - \sin^2 \theta \) (Factoring technique.)
- \( \sin \theta = 2 \cos^2 \theta - 1\) (Using an identity and factoring.)
- \( \cos 3\theta + 1 = 0\) (Dealing with horizontal stretch/shrink in the input.)
- \( \tan \left( \theta + \frac{\pi}{4} \right) + \sin^2 \theta = 1 - \cos^2 \theta \) (Using an identity and dealing with a phase shift.)
Solution
1. First, I bring all the trig stuff together and move everything else over to the other side to clean things up.
\[ \sin^2 \theta + \sin \theta = 1 - 1 = 0 \notag \]
Now, I notice that the two terms on the left have a common factor, so I pull it out.
\[ (\sin \theta)(\sin \theta + 1) = 0 \notag \]
Now, there are two ways that this equation could turn out to be true: either \( \sin \theta = 0\) or \( \sin \theta + 1 = 0\). Either of those cases would cause the whole left side to wipe out. In the first case, \( \sin \theta = 0\) when \(\theta = 0, \pi, 2\pi, ...\) etc. On the other hand, \( \sin \theta = -1\) at \( \theta = \frac{3\pi}{2}, \frac{7\pi}{2}, - \frac{\pi}{2},\) etc. So this equation has solutions \( \theta = k\pi\) and \( \theta = \frac{3\pi}{2} + 2k\pi\), where \(k\) can be any integer.
2. The thing to notice here is the \(\cos^2 \theta\) term, which we could rewrite in terms of \( \sin^2 \theta\) using the Pythagorean Identity. Since \(\cos^2 \theta = 1 - \sin^2 \theta\), we sub in and simplify:
\[ \sin \theta = 2(1-\sin^2 \theta) - 1 \quad \rightarrow \quad \sin \theta = -2 \sin^2 \theta + 1 \notag \]
If we're going to use that factoring trick above, we need to get zero on one side. We manipulate...
\[ 2\sin^2 \theta + \sin \theta - 1 = 0 \quad \rightarrow \quad (2 \sin \theta - 1)(\sin \theta + 1) = 0 \notag \]
Again, we have two cases. We have \( 2\sin \theta - 1 = 0\) if \( \sin \theta = \frac{1}{2}\), which happens at \(\theta = \frac{\pi}{6} + 2k\pi, \theta = \frac{5\pi}{6}+2k\pi\), as we've seen. In the other case, we have \( \sin \theta + 1 = 0\) if \(\sin \theta = -1\), which happens at \(\theta = -\frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2},\) etc. So we have solutions
\[ \theta = \frac{\pi}{6} + 2k\pi, \quad \theta = \frac{5\pi}{6}+2k\pi, \quad \theta = \frac{3\pi}{2} + 2k\pi \notag \]
3. The weirdo here is the \(3\theta\) input for cosine, but all we do is ignore that for now. Pretend it's some other nickname. Right now, we're just looking for angles \(x\) where \( \cos x + 1 = 0\). Isolating the trig function term, that is \( \cos x = -1\). We know that happens if \(x = \pi\) on the unit circle, and again for every angle \(\pi+2k\pi\), where \(k\) is any integer, but that's not our final answer! Time to un-nickname and realize this means we want \(3\theta = \pi + 2k \pi \). Solving for \(\theta\), this tells us that our solution list is \( \theta = \frac{\pi+2k \pi}{3}\), where \(k\) is any integer. You could write that as \( (2k+1)\frac{\pi}{3} \), if you want.
4. It looks like there's a lot going on here, but the \(\sin^2 \) and \(\cos^2\) appearing makes me think I should use the Pythagorean Identity to clean things up first. Simplifying, I get
\[ \tan \left( \theta + \frac{\pi}{4} \right) + \sin^2 \theta + \cos^2 \theta = 1 \quad \rightarrow \quad \tan \left( \theta + \frac{\pi}{4} \right) + 1 = 1 \quad \rightarrow \quad \tan \left( \theta + \frac{\pi}{4} \right) = 0 \notag \]
I start by ignoring the garbage in tangent's input. Just look for angles \(x\) where \( \tan x = 0\). This will happen anytime \( \sin x = 0\), which we've already seen is when \(x = k\pi,\) where \(k\) is any integer. That means we want \( \theta + \frac{\pi}{4} = k\pi\). Solving for \(\theta\), we get the answer \(\theta = k\pi-\frac{\pi}{4}\), where \(k\) is any integer. These are values such as \(\theta = -\frac{\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{11\pi}{4}\), etc.
Solve the equations.
- \( \sin(2\theta) = \frac{\sqrt{3}}{2} \)
- \( \cos(\theta - \pi) + 1 = 0\)
- \( \sin 2\theta - \cos \theta = 0 \)
- \( \cos \theta + 1 = \sin \theta\) (Hints: First square both sides of the equation, then use an identity to write everything in terms of cosine. Then you can use the factoring technique. At the end, you MUST check your answers to make sure they're valid in the original equation.)
- Answer
-
1. \(\theta = \frac{\pi}{6}+k\pi, \theta = \frac{\pi}{3}+k\pi,\) where \(k\) is any integer.
2. \(\theta = 2k\pi\), where \(k\) is any integer.
3. Hint: use a double-angle formula on the first term before anything else. Use factoring on this and find solutions for two cases. All together, \(\theta\) can be any odd multiple of \(\frac{\pi}{2}\), as well as \(\theta = \frac{\pi}{6} + 2k\pi, \theta = \frac{5\pi}{6}+2k\pi\), for any integer \(k\).
4. After squaring both sides, which requires FOILing, you should have \( \cos^2 \theta + 2\cos \theta + 1 = \sin^2 \theta\). The next hint is to make everything in terms of cosine by using the Pythagorean Identity to replace \(\sin^2 \theta\)... After factoring and solving your cases, you should get hypothetical solutions \( \theta = \frac{\pi}{2}, \frac{3\pi}{2}\) and \(\theta = \pi\). Checking these answers by plugging them into the original equation, we see that only \( \frac{\pi}{2}\) and \(\pi\) actually work. So we have \(\theta = \frac{\pi}{2}+2k\pi\), \(\theta = (2k+1)\pi\) (which is odd multiples of \(\pi\)).
More practice can be found in the exercises section! I'll leave you with an example straight outta Calc I where you need these skills.
[SPOILER ALERT] The derivative of a function \(f(x)\) is another function who describes the instantaneous slope of the curve \(y = f(x)\) at any particular location \(x\) (the slope of the tangent line at that point, if you remember that discussion from Section 5.1). It's denoted \(f'(x)\), and you will learn allll about this concept in Calc I. It turns out that for \(f(x) = \sin x\), the derivative is \(f'(x) = \cos x\). Now, when doing optimization problems (problems where you are looking for an "optimal" situation, like maximizing profits, or minimizing costs) you must find certain \(x\)-values known as critical numbers. These are locations at which \(f'(x) = 0\).
So to find the critical numbers for the function \(f(x) = \sin x\), you are looking for \(x\) where \( f'(x) = \cos x = 0\). What are all such critical numbers of \(f\)?
Solution
All I want to do is figure out what \(x\) should be to cause \(\cos x = 0\). I know from the unit circle that cosine is zero if \(x = \frac{\pi}{2}, \frac{3\pi}{2}\). Any \(x = \frac{\pi}{2} + 2k\pi\) or \(x = \frac{3\pi}{2} + 2k \pi\) will therefore be a critical number. Writing out some specific examples, we have
\[x = ..., -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, ... \notag \]
from which we can see the critical numbers are all odd multiples of \(\frac{\pi}{2}\).