Answer
$\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$
Work Step by Step
Step 1. Rewrite the equation to get $1-2sin^2\theta+6sin^2\theta=4\longrightarrow sin^2\theta=\frac{3}{4}\longrightarrow sin\theta=\pm\frac{\sqrt 3}{2}$
Step 2. For $sin\theta=\frac{\sqrt 3}{2}$, we have $\theta=2k\pi+\frac{\pi}{3}$ and $\theta=2k\pi+\frac{2\pi}{3}$
Step 3. For $sin\theta=-\frac{\sqrt 3}{2}$, we have $\theta=2k\pi+\frac{4\pi}{3}$ and $\theta=2k\pi+\frac{5\pi}{3}$
Step 4. Within $[0,2\pi)$, we have $\theta=\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$
