Answer
$\{ \frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2} \}$
Work Step by Step
Step 1. Use the identity $cos^2x=1-sin^2x$, we have $2-2sin^2x-sin(x)-1$ or $2sin^2(x)+sin(x)-1=0$, which gives two solutions $sin(x)=-1$ and $sin(x)=\frac{1}{2}$
Step 2. For $sin(x)=-1$, we can find all x-values in $[0,2\pi)$ as $x=\frac{3\pi}{2}$
Step 3. For $sin(x)=\frac{1}{2}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{6},\frac{5\pi}{6}$
Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{ \frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2} \}$
