## Precalculus (6th Edition)

$x\displaystyle \in\{ \frac{\pi}{3},\frac{4\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}\}$
... dividing with 2, $\displaystyle \cos 2x=-\frac{1}{2}$ If $0 \leq x < 2\pi$ then $0 \leq 2x < 4\pi$ So, $2x\in[0,4\pi)$, meaning that if $2$x (a solution) is any number from $[0,2\pi)$, then $2x+2\pi$ is also a solution. Using the unit circle,$2x$ can be $\displaystyle \frac{2\pi}{3} \ \$and $\displaystyle \frac{2\pi}{3}+2\pi=\frac{8\pi}{3},$ (quadrant II), or $\displaystyle \frac{4\pi}{3} \ \$and $\displaystyle \frac{4\pi}{3}+2\pi=\frac{10\pi}{3}$ (quadrant III) So, $2x\displaystyle \in\{ \frac{2\pi}{3},\frac{8\pi}{3},\frac{4\pi}{3},\frac{10\pi}{3}\}$ x is half of 2x, so the corresponding values for x are half of the above corresponding values, Solution set: $x\displaystyle \in\{ \frac{\pi}{3},\frac{4\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}\}$