Answer
$0, \frac{2\pi}{3},\frac{4\pi}{3} $
Work Step by Step
1. Use the identity $cos(2x)=2cos^2x-1$, we have $cos(x)=2cos^2x-1\longrightarrow 2cos^2x-cos(x)-1$ which gives $cos(x)=1$ and $cos(x)=-\frac{1}{2}$
2. For $cos(x)=1$ with $x\in[0,2\pi)$, we can find $x=0$
3. For $cos(x)=-\frac{1}{2}$ with $x\in[0,2\pi)$, we can find $x=\frac{2\pi}{3},\frac{4\pi}{3} $
