5 Easy Steps for Solving Multiple Angle Trig Equations

I’m teaching Pre-Calculus this semester and one of the topics is trig equations. There are many types of trig equations, but the one that caught my interest the most was the one involving multiple angles. I guess this is because I can’t ever remember formally learning how to solve these, but I do remember coming across these in the past and not really knowing how to handle them, so it was nice to learn about a formal way to solve them. It can be a little confusing though, so I broke the process down into 5 easy steps.

*Note: This post already assumes you have had an introduction to the six trig functions and the unit circle. I will try to be as thorough as I can here, but to keep this post clear and concise, I won’t explain things that should be understood before attempting to solve trig equations.

I will use a sample problem to illustrate the 5 steps for solving multiple angle trig equations.

Sample Problem 1

Let’s consider the following trig equation:

\begin{equation*}
\cos 2\theta = \frac{\sqrt{3}}{2}, \quad 0 \leq \theta < 2\pi \end{equation*}

Step 1

First, write down all of the angles between \(0\) and \(2\pi\) that have a cosine of \(\frac{\sqrt{3}}{2}\). You can consult the unit circle if you need to, but eventually you should be able to do this without extra help.

Unit Circle

We get the angles \(\frac{\pi}{6}\) and \(\frac{11\pi}{6}\).

Step 2

Our original equation will be satisfied when \(2\theta = \frac{\pi}{6}\) or \(2\theta = \frac{11\pi}{6}\). What’s important to note here is that the equation will also be satisfied when \(2\theta\) equals any angle that is coterminal with either of these angles (for example, \(-\frac{\pi}{6}\) is coterminal with \(\frac{11\pi}{6}\)). Since cosine has a period of \(2\pi\), if we add or subtract any integer multiple of \(2\pi\) to \(\frac{\pi}{6}\) or \(\frac{11\pi}{6}\), we will get another angle whose cosine is \(\sqrt{3}/2\).

Step 3

Take your realization from Step 2 and write it out in equation form. Write the following two equations in separate columns.

\begin{equation*}
2\theta = \frac{\pi}{6} + 2n\pi
\end{equation*}
\begin{equation*}
2\theta = \frac{11\pi}{6} + 2n\pi
\end{equation*}

The variable \(n\) can be any integer. Writing the formulas this way captures the angles in between \(0\) and \(2\pi\), as well as all of these ones in the same position as them.

Step 4

Solve for \(\theta\) in the two equations you wrote above.

\begin{equation*}
\theta = \frac{\pi}{12} + n\pi
\end{equation*}
\begin{equation*}
\theta = \frac{11\pi}{12} + n\pi
\end{equation*}

Step 5

Recall that the original problem wanted us to find all solutions in the interval \([0, 2\pi)\). What we need to do now is substitute values for \(n\) in the equations from Step 4 to find all of the solutions in \([0, 2\pi)\). We do this by starting with \(n=0\) and continuing with \(n=1,2,3,…\) until we get values that are larger than \(2\pi\). Here’s what it looks like:

  \begin{equation*}
\theta = \frac{\pi}{12} + n\pi
\end{equation*}
\begin{equation*}
\theta = \frac{11\pi}{12} + n\pi
\end{equation*}
\(n=0\) \(\frac{\pi}{12}\) \(\frac{11\pi}{12}\)
\(n=1\) \(\frac{13\pi}{12}\) \(\frac{23\pi}{12}\)
\(n=2\) \(\frac{25\pi}{12}\) \(\frac{35\pi}{12}\)

At this point, we can stop because the answers corresponding to \(n=2\) are greater than \(2\pi\), so our solution set is \(\left\{\frac{\pi}{12},\frac{11\pi}{12},\frac{13\pi}{12},\frac{23\pi}{12}\right\}\).

What’s fun to note here is the number of solutions we got: four. Normally, a trig equation with \(\sin\theta\) or \(\cos\theta\) will just have two solutions in \([0, 2\pi)\). But here, we had an equation with \(\cos2\theta\) and got four solutions. This is because multiplying \(\theta\) by \(2\) in effect, “speeds up” the function so it hits all the same values twice as many times as it normally does. We will see more example of this in the sample problems below.

Cos(theta)

The colored lines indicate two solutions

cosine(2theta)

The colored lines indicate four solutions


Here are a few more example problems to help you get the idea.

Sample Problem 2

\begin{equation*}
\sin3\theta = -\frac{\sqrt{2}}{2}, \quad 0 \leq \theta < 2\pi \end{equation*}

Step 1

Write down all of the angles between \(0\) and \(2\pi\) that have a sine of \(-\frac{\sqrt{2}}{2}\):

We get the angles \(\frac{5\pi}{4}\) and \(\frac{7\pi}{4}\).

Step 2

Realize that our original equation will be satisfied when \(3\theta = \frac{5\pi}{4}\) or \(3\theta = \frac{7\pi}{4}\). We must add \(2n\pi\) to both of these to capture all of their coterminal angles.

Step 3

Take your realization from Step 2 and write it out in equation form. Write the following two equations in separate columns.

\begin{equation*}
3\theta = \frac{5\pi}{4} + 2n\pi
\end{equation*}
\begin{equation*}
3\theta = \frac{7\pi}{4} + 2n\pi
\end{equation*}

Step 4

Solve for \(\theta\) in the two equations you wrote above.

\begin{equation*}
\theta = \frac{5\pi}{12} + \frac{2n\pi}{3}
\end{equation*}
\begin{equation*}
\theta = \frac{7\pi}{12} + \frac{2n\pi}{3}
\end{equation*}

Step 5

Plug in values of \(n\), starting with \(n=0\), and continuing with \(n=1,2,3,…\) until we get values that are larger than \(2\pi\). I find that it helps to write the expressions from the previous step with the same denominator. Here’s what it looks like:

  \begin{equation*}
\theta = \frac{5\pi}{12} + \frac{8n\pi}{12}
\end{equation*}
\begin{equation*}
\theta = \frac{7\pi}{12} + \frac{8n\pi}{12}
\end{equation*}
\(n=0\) \(\frac{5\pi}{12}\) \(\frac{7\pi}{12}\)
\(n=1\) \(\frac{13\pi}{12}\) \(\frac{15\pi}{12}\)
\(n=2\) \(\frac{21\pi}{12}\) \(\frac{23\pi}{12}\)
\(n=3\) \(\frac{29\pi}{12}\) \(\frac{31\pi}{12}\)

The last row is outside of the specified range. Our solution set (in lowest terms) is \(\left\{\frac{5\pi}{12},\frac{7\pi}{12},\frac{13\pi}{12},\frac{5\pi}{4}, \frac{21\pi}{12}, \frac{23\pi}{12}\right\}\).

Sample Problem 3

\begin{equation*}
\tan2\theta = -\frac{\sqrt{3}}{3}, \quad 0 \leq \theta < 2\pi \end{equation*}

Step 1

Trig equations involving tangent and cotangent are solved the same way as those with the other trig functions, except you have to remember that the period of tangent and cotangent is \(\pi\). This changes things a little bit.

Since tangent has a period of \(\pi\), we only need to write down all of the angles between \(0\) and \(\pi\) that have a tangent of \(-\frac{\sqrt{3}}{3}\):

We get the angle \(\frac{2\pi}{3}\).

Step 2

Once more, our original equation will be satisfied when \(2\theta = \frac{2\pi}{3}\). Since we are dealing with tangent, we must add \(n\pi\) (rather than \(2n\pi\)) to capture all of the angles with a tangent of \(-\frac{\sqrt{3}}{3}\).

Step 3

Take your realization from Step 2 and write it out in equation form.

\begin{equation*}
2\theta = \frac{2\pi}{3} + n\pi
\end{equation*}

Step 4

Solve for \(\theta\) in the equation you wrote above.

\begin{equation*}
\theta = \frac{2\pi}{6} + \frac{n\pi}{2}
\end{equation*}

Step 5

Plug in values of \(n\), starting with \(n=0\), and continuing with \(n=1,2,3,…\) until we get values that are larger than \(2\pi\). I find that it helps to write the expressions from the previous step with the same denominator. Here’s what it looks like:

  \begin{equation*}
\theta = \frac{2\pi}{6} + \frac{3n\pi}{6}
\end{equation*}
\(n=0\) \(\frac{2\pi}{6}\)
\(n=1\) \(\frac{5\pi}{6}\)
\(n=2\) \(\frac{8\pi}{6}\)
\(n=3\) \(\frac{11\pi}{6}\)

Any further calculation will give solutions larger than \(2\pi\), so our solution set (in lowest terms) is \(\left\{\frac{\pi}{3},\frac{5\pi}{6},\frac{4\pi}{3},\frac{11\pi}{6}\right\}\).


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