Answer
$\{\frac{\pi}{6},\frac{5\pi}{6}, \frac{3\pi}{2} \}$
Work Step by Step
Step 1. Using the identity $cos(2x)=1-2sin^2(x)$, we have $2sin^2(x)+sin(x)-1=0$ or $(sin(x)+1)(2sin(x)-1)=0$, which gives $sin(x)=-1$ 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 $\{\frac{\pi}{6},\frac{5\pi}{6}, \frac{3\pi}{2} \}$
