## Precalculus (6th Edition) Blitzer

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