Answer
$\frac{\pi}{6}$, $\frac{\pi}{2}$, $\frac{3\pi}{2}$, $\frac{11\pi}{6}$
Work Step by Step
$2\cos^2 x-\sqrt{3}\cos x=0$
$\cos x(2\cos x-\sqrt{3})=0$
$\cos x=0$ or $2\cos x-\sqrt{3}=0$
If $\cos x=0$, the only solutions in $[0, 2\pi)$ are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.
If $2\cos x-\sqrt{3}=0$:
$2\cos x=\sqrt{3}$
$\cos x=\frac{\sqrt{3}}{2}$
The only solutions in $[0, 2\pi)$ are $\frac{\pi}{6}$ and $\frac{11\pi}{6}$.
