Answer
$\theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}$
Work Step by Step
Since $ sin2\theta=2sin\theta\cdot cos\theta $, we can rewrite the equation as
$2sin\theta\cdot cos\theta-cos\theta=0$, or $ cos\theta(2sin\theta-1)=0$.
The solutions are $ cos\theta=0$ or $2sin\theta-1=0$ which is $ sin\theta=\frac{1}{2}$
Within $[0,2\pi]$, we have $\theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}$
