Answer
$0,\frac{\pi}{5},\frac{2\pi}{5},\frac{3\pi}{5},\frac{4\pi}{5},\pi,\frac{6\pi}{5},\frac{7\pi}{5},\frac{8\pi}{5},\frac{9\pi}{5}$
Work Step by Step
1. Use the Sum-to-Product Formula, we have:
$cos(4\theta)-cos(6\theta)=0\Longrightarrow
-2sin(\frac{4\theta+6\theta}{2})sin(\frac{4\theta-6\theta}{2})=0\Longrightarrow sin(5\theta)sin(\theta)=0$
2. For $sin(\theta)=0$, we have $\theta=k\pi$
3. For $sin(5\theta)=0$, we have $5\theta=k\pi$, thus $\theta=\frac{k\pi}{5},$
4. Within $[0,2\pi)$, we have $\theta=0,\frac{\pi}{5},\frac{2\pi}{5},\frac{3\pi}{5},\frac{4\pi}{5},\pi,\frac{6\pi}{5},\frac{7\pi}{5},\frac{8\pi}{5},\frac{9\pi}{5}$
