Answer
$\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$
Work Step by Step
Step 1. Rewrite the equation $1+sin\theta=2(1-sin^2\theta) \longrightarrow 2sin^2\theta+sin\theta-1=0 \longrightarrow (2sin\theta-1)(sin\theta+1)=0$. Thus $sin\theta=\frac{1}{2}$ or $sin\theta=-1$.
Step 2. For $sin\theta=\frac{1}{2}$, we have $\theta=\frac{\pi}{6}, \frac{5\pi}{6}$
Step 3. For $sin\theta=-1$, we have $\theta=\frac{3\pi}{2}$
