Solving Equations Involving Trig Functions

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*}

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To find solutions to this equation, we find the radian value(s) that will give us sine equal to \(\frac{1}{2}\text{.}\)

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For \(x\) in \([0,2\pi]\text{,}\) we have two solutions: \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\text{.}\)

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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.

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Example 5.5.2.

Solve \(2\sin 2x = -1\) for \(x\) in \([0,2\pi]\text{.}\)

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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 \(-\frac{1}{2}\text{?}\)

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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*}

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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*}

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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*}

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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.

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Example 5.5.3.

Solve \(2\cos^2x\tan x - \tan x = 0\) for \(x\) in \([0,2\pi]\text{.}\)

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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*}

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If \(\tan x = 0\text{,}\) then \(x=0,\pi, 2\pi\text{.}\)

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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 \([0,2\pi]\text{,}\) \(x=\frac{pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\text{.}\)

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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*}

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In some problems we will first use identities and then factor.

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Example 5.5.4.

Solve \(\cos x = \cos 2x\) for \(x\) in \([0,2\pi]\text{.}\)

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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 \(\cos x - 1 =0\text{.}\)

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If \(2\cos x + 1 = 0\) then \(\cos x = -\frac{1}{2}\) and \(x= \frac{2\pi}{3}, \frac{4\pi}{3}\text{.}\)

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If \(\cos x - 1 =0\) then \(\cos x =1\) and \(x = 0, 2\pi\text{.}\)

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So the solution set is

\begin{equation*} \left\{0, 2\pi, \frac{2\pi}{3}, \frac{4\pi}{3} \right\}. \end{equation*}

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