## Trigonometry (11th Edition) Clone

The solution set is $$\{60^\circ, 120^\circ\}$$
$$2\sin\theta-\sqrt3=0$$ over the interval $[0^\circ,360^\circ)$ We carry out normal algebra here: $$\sin\theta=\frac{\sqrt3}{2}$$ $$\theta=\sin^{-1}\frac{\sqrt3}{2}$$ Over the interval $[0^\circ,360^\circ)$, there are two values of $\theta$ whose $\sin\theta=\frac{\sqrt3}{2}$, which are $60^\circ$ and $120^\circ$. In other words, $$\theta=\{60^\circ, 120^\circ\}$$ That is also the solution set to the problem.