Answer
$\{\frac{\pi}{2},\frac{2\pi}{3},\frac{4\pi}{3},\frac{3\pi}{2}\}$
Work Step by Step
Step 1. Using the identity $cos(2x)=2cos^2(x)-1$, we have $2cos^2(x)+cos(x)=0$ or $cos(x)(2cos(x)+1)=0$, which gives $cos(x)=0$ and $cos(x)=-\frac{1}{2}$
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 $cos(x)=-\frac{1}{2}$ , 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 $\{\frac{\pi}{2},\frac{2\pi}{3},\frac{4\pi}{3},\frac{3\pi}{2}\}$
