Answer
The solution set is $$\{60^\circ, 120^\circ\}$$
Work Step by Step
$$2\sin\theta-\sqrt3=0$$ 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.
