Answer
$\displaystyle \{\frac{\pi}{12},\frac{13\pi}{12},\frac{11\pi}{12},\frac{23\pi}{12}\}$
Work Step by Step
... dividing with 2,
$\displaystyle \cos 2x=\frac{\sqrt{3}}{2}$
If $ 0 \leq x < 2\pi$ then
$ 0 \leq 2x < 4\pi$
So, $2x\in[0,4\pi)$, meaning that
if 2x (a solution) is any number from $[0,2\pi)$, then
$ 2x+2\pi$ is also a solution.
Using the unit circle,$ 2x$ can be
$\displaystyle \frac{\pi}{6}$ or $\displaystyle \frac{\pi}{6}+2\pi=\frac{13\pi}{6},$ (quadrant I),
or
$\displaystyle \frac{11\pi}{6}$ or $\displaystyle \frac{11\pi}{6}+2\pi=\frac{23\pi}{6}$ (quadrant IV)
So, $2x\displaystyle \in\{ \frac{\pi}{6},\frac{13\pi}{6},\frac{11\pi}{6},\frac{23\pi}{6}\}$
x is half of 2x, so the corresponding values for x are
half of the above values,
$\displaystyle \frac{\pi}{6}\div 2=\frac{\pi}{12},$
$\displaystyle \frac{13\pi}{6}\div 2=\frac{13\pi}{12},$
$\displaystyle \frac{11\pi}{6}\div 2=\frac{11\pi}{12},$
$\displaystyle \frac{23\pi}{6}\div 2=\frac{23\pi}{12}.$
Solution set:
$x\in\displaystyle \{\frac{\pi}{12},\frac{13\pi}{12},\frac{11\pi}{12},\frac{23\pi}{12}\}$
