Answer
$\{ \frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2} \}$
Work Step by Step
Step 1. Factor the equation as $cos(x)(1-2sin(x))=0)$, which gives two solutions $cos(x)=0$ and $sin(x)=\frac{1}{2}$
Step 2. For $cos(x)=0$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{2},\frac{3\pi}{2}$
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 $\{ \frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2} \}$
