Section 5.5 Solving Equations Involving Trig Functions
We will begin by considering equations with one term involving a trigonometric function.
Example 5.5.1. An equation with one trigonometric term.
Solve \(2\sin x = 1\text{.}\)
We want to find which values of \(x\) make this equation true. We will not rewrite this equation as \(x=\cdots\text{.}\) We will isolate the trig function instead:
\begin{equation*}
\sin x = \frac{1}{2}
\end{equation*}
To find solutions to this equation, we find the radian value(s) that will give us sine equal to \(\frac{1}{2}\text{.}\)
For \(x\) in \([0,2\pi]\text{,}\) we have two solutions: \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\text{.}\)
However, if \(x\) can be any number, note that \(\frac{\pi}{6}+2\pi\text{,}\) \(\frac{\pi}{6}+4\pi\text{,}\) \(\frac{\pi}{6}-2\pi\text{,}\) and \(\frac{5\pi}{6}+2\pi\) are also solutions. In fact, for \(x\) in \((-\infty,\infty)\text{,}\) we have an infinite number of solutions that can be represented as
\begin{align*}
x \amp = \frac{\pi}{6} \amp \amp \text{ and } \amp x \amp = \frac{5\pi}{6}+2n\pi
\end{align*}
where \(n\) is any integer.
Example 5.5.2.
Solve \(2\sin 2x = -1\) for \(x\) in \([0,2\pi]\text{.}\)
Solution.
We will begin by solving for \(\sin 2x\text{.}\)
\begin{equation*}
\sin 2x = -\frac{1}{2}
\end{equation*}
We want solutions \(x\) for which
\begin{equation*}
0 \leq x \leq 2\pi \text{ or } 0 \leq x \leq 4\pi.
\end{equation*}
For which radian values \(\theta\) between \(0\) and \(4\pi\) does \(\sin \theta \) equal \(-\frac{1}{2}\text{?}\)
Between \(0\) and \(2\pi\text{,}\) \(\sin \theta = -\frac{1}{2}\) for
\begin{equation*}
\theta = \frac{7\pi}{6} \text{ and } \theta = \frac{11\pi}{6}.
\end{equation*}
Between \(2\pi\)and \(4\pi\text{,}\) \(\sin\theta=-\frac{1}{2}\) for
\begin{equation*}
\theta=\frac{7\pi}{6}+2\pi = \frac{19\pi}{6} \text{ and } \theta = \frac{11\pi}{6}+2\pi=\frac{23\pi}{6}.
\end{equation*}
To compensate for the \(2x\) in \(\sin 2x\text{,}\) we will set \(\theta=2x\text{.}\)
\begin{equation*}
\sin \frac{7\pi}{6} = -\frac{1}{2} = \sin 2x \Rightarrow 2x =\frac{7\pi}{6} \Rightarrow x=\frac{7\pi}{12}
\end{equation*}
Similarly,
\begin{align*}
\sin \frac{11\pi}{6} \amp -\frac{1}{2} = \sin 2x \amp \amp \Rightarrow \amp x \amp = \frac{11\pi}{12}, \\
\sin \frac{19\pi}{6} \amp -\frac{1}{2} = \sin 2x \amp \amp \Rightarrow \amp x \amp = \frac{19\pi}{12},
\end{align*}
and
\begin{align*}
\sin \frac{23\pi}{6} \amp -\frac{1}{2} = \sin 2x \amp \amp \Rightarrow \amp x \amp = \frac{23\pi}{12}.
\end{align*}
Thus the solutions for \(x\) in \([0,2\pi]\) are
\begin{equation*}
x = \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{19\pi}{12}, \frac{23\pi}{12}.
\end{equation*}
We will now consider equations with more than one term involving a trigonometric function. The key concept involves in these types of equations is factoring.
Example 5.5.3.
Solve \(2\cos^2x\tan x - \tan x = 0\) for \(x\) in \([0,2\pi]\text{.}\)
Solution.
We begin by factoring \(\tan x\) out of each term.
\begin{equation*}
\tan x \left(2\cos^2 x - 1 \right) = 0.
\end{equation*}
Thus either
\begin{align*}
\tan x \amp = 0 \amp \amp or \amp 2 \cos^2x - 1 \amp = 0
\end{align*}
If \(\tan x = 0\text{,}\) then \(x=0,\pi, 2\pi\text{.}\)
If \(2\cos^2 x - 1 = 0\text{,}\) then \(\cos^2x=\frac{1}{2}\) and \(\cos x = \pm \frac{1}{\sqrt{2}} = \mp \frac{\sqrt{2}}{2}\text{.}\) So for \(x\) in \([0,2\pi]\text{,}\) \(x=\frac{pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\text{.}\)
The solution set is
\begin{equation*}
\left\{0, \pi, 2\pi, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\}.
\end{equation*}
In some problems we will first use identities and then factor.
Example 5.5.4.
Solve \(\cos x = \cos 2x\) for \(x\) in \([0,2\pi]\text{.}\)
Solution.
We will first use the identity \(\cos2x = 2\cos^2 x - 1\text{.}\)
\begin{align*}
\cos x \amp = 2\cos^2x - 1 \\
2\cos^2x - 1 - \cos x \amp = 0
\end{align*}
Next, factor.
\begin{align*}
2\cos^2x - \cos x - 1\amp = 0 \\
\left(2\cos x + 1\right)\left(\cos x - 1\right)\amp = 0
\end{align*}
Thus either \(2\cos x + 1 = 0\) or \(\cos x - 1 =0\text{.}\)
If \(2\cos x + 1 = 0\) then \(\cos x = -\frac{1}{2}\) and \(x= \frac{2\pi}{3}, \frac{4\pi}{3}\text{.}\)
If \(\cos x - 1 =0\) then \(\cos x =1\) and \(x = 0, 2\pi\text{.}\)
So the solution set is
\begin{equation*}
\left\{0, 2\pi, \frac{2\pi}{3}, \frac{4\pi}{3} \right\}.
\end{equation*}
