## Precalculus (6th Edition) Blitzer

$\pi$
Step 1. Let $u=cos(x)$; we have $u^2-2u-3=(u-3)(u+1)=0$ which gives $u=-1, 3$ Step 2. For $cos(x)=u=-1$, we have solutions in $[0,2\pi)$ as $x=\pi$ Step 3. For $cos(x)=u=3$, we have no solutions in $[0,2\pi)$. Step 4. Thus, we have all the solutions in $[0,2\pi)$ as $x=\pi$