Answer
$\frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, $\frac{7\pi}{4}$
Work Step by Step
$4\sin^2 x+2\cos^2 x=3$
$2\sin^2 x+2\sin^2 x+2\cos^2 x=3$
$2\sin^2 x+2(\sin^2 x+\cos^2 x)=3$
$2 \sin^2 x+2=3$
$2\sin^2 x=1$
$\sin^2 x=\frac{1}{2}$
$\sin x=\pm\frac{\sqrt{2}}{2}$
If $\sin x=\frac{\sqrt{2}}{2}$, then the only solutions in $[0, 2\pi)$ are $\frac{\pi}{4}$ and $\frac{3\pi}{4}$.
If $\sin x=-\frac{\sqrt{2}}{2}$, then the only solutions in $[0, 2\pi)$ are $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$.
