Answer
$\frac{\pi}{6}$, $\frac{5\pi}{6}$
Work Step by Step
$2\sin^2 x-5\sin x+2=0$
$(2\sin x-1)(\sin x-2)=0$
If $2\sin x-1=0$, then $2\sin x=1$, and $\sin x=\frac{1}{2}$. The only solutions in $[0, 2\pi)$ are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$.
if $\sin x-2=0$, then $\sin x=2$, which is impossible because $\sin x$ cannot have a value greater than 1.
