Answer
$$
\theta=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \frac{11 \pi}{6}
$$
Work Step by Step
Since $$\cos \alpha=-\cos \beta$$
when $\alpha+\beta=\pi+2 \pi k$ or $\alpha=\beta+\pi+2 \pi k .$
Substituting $\alpha=4 \theta$ and $\beta=2 \theta,$ we have
either
$$6 \theta=\pi+2 \pi k$$
or
$$4 \theta=2 \theta+\pi+2 \pi k .$$
Solving each of these equations for $\theta$ yields $$\theta=\frac{\pi}{6}+\frac{\pi}{3} k$$ or $$\theta=\frac{\pi}{2}+\pi k$$
The solutions on the interval $0 \leq \theta<2 \pi$ are then
$$
\theta=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \frac{11 \pi}{6}
$$
