Question

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

$\cos \left(2\theta \right)=-\frac{1}{2}$

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$0\le \theta \le 2\pi$

$\cos \left(2\theta \right)=-\frac{1}{2}$

Step 2. In the interval [0,2π), the cosine function equals -12 at 2π3.

So, we know that $2\theta$must equal $\frac{2\mathrm{\pi }}{3}$.

To find these solutions, write the general formula that gives all the solutions.

localid="1646587269475" $2\theta =\frac{2\mathrm{\pi }}{3}+2\mathrm{\pi n}$

Divide by 2 on both side

localid="1646592113006" $$\begin{gathered} \frac{2}{2}\theta =\frac{1}{2}\left(\frac{2\mathrm{\pi }}{3}+2\mathrm{\pi n}\right) \\ \theta =\frac{1}{2}\times \frac{2\mathrm{\pi }}{3}+2\mathrm{\pi n}\times \frac{1}{2} \\ \theta =\frac{\mathrm{\pi }}{3}+\mathrm{\pi n} \end{gathered}$$

$\theta =\frac{2\mathrm{\pi }}{3}+\mathrm{\pi n}$

Step 3. The general formula is θ=π3+πn,θ=2π3+πn

So the value of given function in interval$[0,2\mathrm{\pi })$is

localid="1646592541288" $$\begin{gathered} \theta =\frac{\mathrm{\pi }}{3}+\pi \times 0\theta =\frac{\mathrm{\pi }}{3}+\pi \times 1\theta =\frac{2\mathrm{\pi }}{3}+\pi \times 0\theta =\frac{2\mathrm{\pi }}{3}+\pi \times 1 \\ \theta =\frac{\mathrm{\pi }}{3}\theta =\frac{\mathrm{\pi }}{3}+\pi \theta =\frac{2\mathrm{\pi }}{3}\theta =\frac{2\mathrm{\pi }}{3}+\pi \\ \theta =\frac{\mathrm{\pi }}{3}\theta =\frac{\mathrm{\pi }+3\mathrm{\pi }}{3}\theta =\frac{2\mathrm{\pi }}{3}\theta =\frac{2\mathrm{\pi }+3\mathrm{\pi }}{3} \\ \theta =\frac{\mathrm{\pi }}{3}\theta =\frac{4\mathrm{\pi }}{3}\theta =\frac{2\mathrm{\pi }}{3}\theta =\frac{5\mathrm{\pi }}{3} \end{gathered}$$

So the solution set is localid="1646592566280" $\left\{\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$.