Answer
$\frac{2\pi}{3}$, $\frac{4\pi}{3}$
Work Step by Step
$2\cos ^2\theta+5\cos \theta+2=0$
$(2\cos \theta+1)(\cos \theta+2)=0$
If $2\cos \theta+1=0$, then $2\cos \theta=-1$, and $\cos \theta=-\frac{1}{2}$. The only solutions in $[0, 2\pi)$ are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.
If $\cos \theta+2=0$, then $\cos \theta=-2$. This is impossible because $\cos \theta$ cannot be less than -1.
