## Thomas' Calculus 13th Edition

$\theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}$
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 $\theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}$