Answer
$\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
Work Step by Step
1. $4sin^2\theta-3=0\Longrightarrow sin\theta=\pm\frac{\sqrt 3}{2}$
2. For $sin\theta=\frac{\sqrt 3}{2}$, we have $\theta=2k\pi+\frac{\pi}{3}$ or $\theta=2k\pi+\frac{2\pi}{3}$
3. For $sin\theta=-\frac{\sqrt 3}{2}$, we have $\theta=2k\pi+\frac{4\pi}{3}$ or $\theta=2k\pi+\frac{5\pi}{3}$
4. Within $[0,2\pi)$, we have $\theta=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$