Answer
$\frac{\pi}{3}$, $\frac{2\pi}{3}$, $\frac{4\pi}{3}$, $\frac{5\pi}{3}$
Work Step by Step
$2\cos^2 x+\cos 2x=0$
$2\cos^2 x+2\cos^2 x-1=0$
$4\cos^2 x-1=0$
$(2\cos x+1)(2\cos x-1)=0$
If $2\cos x+1=0$, then $2\cos x=-1$, and $\cos x=-\frac{1}{2}$. The only solutions in $[0, 2\pi)$ are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.
If $2\cos x-1=0$, then $2\cos x=1$, and $\cos x=\frac{1}{2}$. The only solutions in $[0, 2\pi)$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.
