Answer
$\frac{\pi}{2},\frac{3\pi}{2}, \frac{\pi}{6},\frac{5\pi}{6}, \frac{7\pi}{6},\frac{11\pi}{6}$
Work Step by Step
1. $cos(2\theta)+cos(4\theta)=0 \Longrightarrow cos(2\theta)+2cos^2(2\theta)-1=0\Longrightarrow (2cos(2\theta)-1)(cos(2\theta)+1)=0 \Longrightarrow cos2\theta=-1, \frac{1}{2}$
2. For $cos2\theta=-1$, we have $2\theta=2k\pi+\pi$, thus $\theta=k\pi+\frac{\pi}{2}$.
3. For $cos2\theta=\frac{1}{2}$, we have $2\theta=2k\pi+\frac{\pi}{3}$ or $2k\pi+\frac{5\pi}{3}$, thus $\theta=k\pi+\frac{\pi}{6}$ or $k\pi+\frac{5\pi}{6}$,
4. Within $[0,2\pi)$, we have $\theta=\frac{\pi}{2},\frac{3\pi}{2}, \frac{\pi}{6},\frac{5\pi}{6}, \frac{7\pi}{6},\frac{11\pi}{6}$
