Answer
$0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}$.
Work Step by Step
Step 1. Rewrite the equation as $(1-2sin^2x)-sin(x)=1$ or $2sin^2x+sin(x)=sin(x)(2sin(x)+1)=0$
Step 2. For $sin(x)=0$, we have solutions in $[0,2\pi)$ as $x=0,\pi$
Step 3. For $sin(x)=-\frac{1}{2}$, we have solutions in $[0,2\pi)$ as $x=\frac{7\pi}{6}, \frac{11\pi}{6}$
Step 4. Thus, we have all the solutions in $[0,2\pi)$ as $x=0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}$.
