Answer
The solutions of the given equation are $$\{\frac{\pi}{12},\frac{11\pi}{12},\frac{13\pi}{12},\frac{23\pi}{12}\}$$
Work Step by Step
$$\cos2x=\frac{\sqrt3}{2}$$ over interval $[0,2\pi)$
1) Interval $[0,2\pi)$ can be written as
$$0\le x\lt2\pi$$
As a result, for $2x$, the interval would be
$$0\le2x\lt4\pi$$
or $$2x\in[0,4\pi)$$
2) Now consider back the equation $$\cos2x=\frac{\sqrt3}{2}$$
Over the interval $[0,4\pi)$, there are 4 values with $\cos$ equaling $\frac{\sqrt{3}}{2}$, which are $\frac{\pi}{6},\frac{11\pi}{6},\frac{13\pi}{6},\frac{23\pi}{6}$, meaning that
$$2x=\{\frac{\pi}{6},\frac{11\pi}{6},\frac{13\pi}{6},\frac{23\pi}{6}\}$$
So $$x=\{\frac{\pi}{12},\frac{11\pi}{12},\frac{13\pi}{12},\frac{23\pi}{12}\}$$
