Question

In Problems 11–34, solve each equation on the interval $0\le \theta \le 2\pi$.

$4{\cos }^2\theta =1$

Short answer

The solution set is$\left\{\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3}\frac{4\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$

Step-by-step solution

Step 1. Given Information

In the given problem we have to solve each equation on the interval$0\le \theta \le 2\pi$.

$4{\cos }^2\theta =1$

Step 2. Firstly solving the equation4cos2θ=1

Divide by 4 on both side

$$\begin{gathered} \frac{4}{4}{\cos }^2\theta =\frac{1}{4} \\ {\cos }^2\theta =\frac{1}{4} \end{gathered}$$

Taking square root on both side

role="math" localid="1646566212432" $$\begin{gathered} \cos \theta =\pm \sqrt{\frac{1}{4}} \\ \cos \theta =\pm \frac{1}{2} \end{gathered}$$

$\cos \theta =\frac{1}{2}\mathrm{and}\cos \theta =-\frac{1}{2}$

Step 3. After solving the equationcosθ=12 and cosθ=-12

The period of the cosine function $\cos \theta =\frac{1}{2}$is $2\mathrm{\pi }$. In the interval $[0,2\mathrm{\pi })$, there are two angles $\theta$for which

$\cos \theta =\frac{1}{2}:\theta =\frac{\pi }{3}\mathrm{and}\theta =\frac{5\pi }{3}$

Step 4. The period of the cosine function cosθ=-12 is 2π.

In the interval $[0,2\mathrm{\pi })$, there are two angles $\theta$for which $\cos \theta =-\frac{1}{2}:\theta =\frac{2\pi }{3}\mathrm{and}\theta =\frac{4\pi }{3}$

So the solution set is$\left\{\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3}\frac{4\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$