## Precalculus (6th Edition) Blitzer

$0,\frac{\pi}{3}, \pi, \frac{5\pi}{3}$
Step 1. Given the equation $tan(x)=2cos(x)tan(x)$ we have $tan(x)(2cos(x)-1)=0)$, Step 2. For $tan(x)=0$, we can find the solutions in $[0,2\pi)$ as $x=0,\pi$ Step 3. For $cos(x)=\frac{1}{2}$, we can find the solutions in $[0,2\pi)$ as $x=\frac{\pi}{3}, \frac{5\pi}{3}$ Step 4. Thus, we can find all the solutions in $[0,2\pi)$ as $0,\frac{\pi}{3}, \pi, \frac{5\pi}{3}$