#### Answer

$0$, $\frac{2\pi}{3}$, $\frac{4\pi}{3}$

#### Work Step by Step

$2\cos^2 x-\cos x=1$
$2\cos^2 x-\cos x-1=0$
$(2\cos x+1)(\cos x-1)=0$
$2\cos x+1=0$ or $\cos x-1=0$
If $2\cos x+1=0$:
$2\cos x=-1$
$\cos x=-\frac{1}{2}$
The only solutions in $[0, 2\pi)$ are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.
If $\cos x-1=0$:
$\cos x=1$
The only solution in $[0, 2\pi)$ is $0$.