Answer
$\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}$
Work Step by Step
Step 1. Rewrite the equation $1-sin^2\theta-sin^2\theta+sin\theta=0 \longrightarrow 2sin^2\theta-sin\theta-1=0 \longrightarrow (sin\theta-1)(2sin\theta+1)=0$. Thus $sin\theta=1$ or $sin\theta=-\frac{1}{2}$
Step 2. For $sin\theta=1$, we have $\theta=\frac{\pi}{2}$
Step 3. For $sin\theta=-\frac{1}{2}$, we have $\theta=\frac{7\pi}{6},\frac{11\pi}{6}$
