## Precalculus (6th Edition) Blitzer

$\frac{\pi}{2},\frac{3\pi}{2}$
Step 1. Given the equation $cos(2x)=-1$, we have to find the solution in $[0,2\pi)$ as $2x=\pi$ and $x=\frac{\pi}{2}$ Step 2. Consider the original function has a period of $\pi$; we can express all the solutions as $x=k\pi+\frac{\pi}{2}$ where $k$ is an integer. Thus, within $[0,2\pi)$, we have $x=\frac{\pi}{2},\frac{3\pi}{2}$ as solutions to the original equation.