## Precalculus (6th Edition)

$\theta\in\{0^{o},120^{o}, 240^{o}\}$
$\sin 3\theta=0$ If $0^{o} \leq \theta < 360^{o}$ then $0^{o} \leq 3\theta < 1080^{0}$ So, $3\theta\in[0,1080^{o})$, meaning that if $3\theta$ (a solution) is any angle from $[0^{o},360^{o})$, then $(3\theta+360^{o})$ and $(3\theta+720^{o})$ are also solutions. Using the unit circle,$3\theta$ can be $0^{o}$, but also $0^{o}+360^{o}=360^{o}$, and $0^{o}+720^{o}=720^{0}$ So, $3\theta\in\{ 0^{o},360^{o},720^{o}\}$, and, since $(3\theta)\div 3=\theta,$ after dividing each corresponding angle with 3, $\theta\in\{0^{o},120^{o}, 240^{o}\}$