Answer
$x\displaystyle \in\{ \frac{\pi}{3},\frac{4\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}\}$
Work Step by Step
... 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:
$x\displaystyle \in\{ \frac{\pi}{3},\frac{4\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}\}$
