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