Answer
$\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4} $
Work Step by Step
Using trigonometric identities, we have
$2sin^2\theta-1=0 \longrightarrow sin\theta=\pm\frac{\sqrt 2}{2} \longrightarrow \theta=\frac{\pi}{4}, \pi-\frac{\pi}{4},\pi+\frac{\pi}{4},2\pi-\frac{\pi}{4} \longrightarrow \theta=\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4} $
