## Trigonometry (11th Edition) Clone

The solution set is $$\{0^\circ, 60^\circ,120^\circ, 180^\circ, 240^\circ, 300^\circ\}$$
$$\sin3\theta=0$$ over interval $[0^\circ,360^\circ)$ 1) Find corresponding interval for $3\theta$ The interval for $\theta$ is $[0^\circ,360^\circ)$, which can also be written as the inequality: $$0^\circ\le\theta\lt360^\circ$$ Therefore, for $3\theta$, the inequality would be $$0^\circ\le3\theta\lt1080^\circ$$ Thus, the corresponding interval for $3\theta$ is $[0^\circ,1080^\circ)$. 2) Now we examine the equation: $$\sin3\theta=0$$ Over interval $[0^\circ,1080^\circ)$, there are 6 values whose sine equals $0$, which are $\{0^\circ, 180^\circ, 360^\circ,540^\circ, 720^\circ, 900^\circ\}$ Therefore, $$3\theta=\{0^\circ, 180^\circ, 360^\circ,540^\circ, 720^\circ, 900^\circ\}$$ It follows that $$\theta=\{0^\circ, 60^\circ,120^\circ, 180^\circ, 240^\circ, 300^\circ\}$$ This is the solution set of the equation.