Answer
$\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$.
Work Step by Step
Step 1. Rewrite the equation as $2(1-sin^2x)-sin(x)=1$ or $2sin^2(x)+sin(x)-1=0$
Step 2. Let $u=sin(x)$; we have $2u^2+u-1=(2u-1)(u+1)=0$ which gives $u=-1, \frac{1}{2}$
Step 3. For $sin(x)=u=-1$, we have solutions in $[0,2\pi)$ as $x=\frac{3\pi}{2}$
Step 4. For $sin(x)=u=\frac{1}{2}$, we have solutions in $[0,2\pi)$ as $x=\frac{\pi}{6}, \frac{5\pi}{6}$.
Step 5. Thus, we have all the solutions in $[0,2\pi)$ as $x=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$.
