Answer
$\frac{\pi}{3}+k\pi$, $\frac{2\pi}{3}+k\pi$
Work Step by Step
$4\sin^2\theta-3=0$
$4\sin^2\theta=3$
$\sin^2\theta=\frac{3}{4}$
$\sin\theta=\pm\frac{\sqrt{3}}{2}$
If $\sin\theta=\frac{\sqrt{3}}{2}$, then $\theta=\frac{\pi}{3}+2k\pi$ or $\theta=\frac{2\pi}{3}+2k\pi$.
If $\sin\theta=-\frac{\sqrt{3}}{2}$, then $\theta=\frac{4\pi}{3}+2k\pi$ or $\theta=\frac{5\pi}{3}+2k\pi$.
Combining the two cases, we get $\theta=\frac{\pi}{3}+k\pi$ or $\theta=\frac{2\pi}{3}+k\pi$.
