Answer
$\frac{\pi}{3}$, $\frac{5\pi}{3}$
Work Step by Step
$2\cos^2x-7\cos x+3=0$
$(2\cos x-1)(\cos x-3)=0$
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 $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.
If $\cos x-3=0$, then $\cos x=3$, which is impossible because $\cos x$ cannot be greater than 1.
