## Calculus (3rd Edition)

$$\theta=0, \frac{2 \pi}{5}, \frac{4 \pi}{5}, \pi, \frac{6 \pi}{5}, \frac{8 \pi}{5}$$
Since $$\sin \alpha=-\sin \beta$$ when $\alpha=-\beta+2 \pi k$ or $\alpha=\pi+\beta+2 \pi k .$ Substituting $\alpha=2 \theta$ and $\beta=3 \theta,$ we have either $$2 \theta=-3 \theta+2 \pi k$$ or $$2 \theta=\pi+3 \theta+2 \pi k .$$ Solving each of these equations for $\theta$ yields $\theta=\frac{2}{5} \pi k$ or $\theta=-\pi-2 \pi k$. The solutions on the interval $0 \leq \theta<2 \pi$ are then $$\theta=0, \frac{2 \pi}{5}, \frac{4 \pi}{5}, \pi, \frac{6 \pi}{5}, \frac{8 \pi}{5}$$