Answer
The solution set is $$\{15^\circ, 75^\circ, 195^\circ,255^\circ\}$$
Work Step by Step
$$2\sqrt3\sin2\theta=\sqrt3$$ over interval $[0^\circ,360^\circ)$
1) Find corresponding interval for $2\theta$
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 $2\theta$, the inequality would be
$$0^\circ\le2\theta\lt720^\circ$$
Thus, the corresponding interval for $2\theta$ is $[0^\circ,720^\circ)$.
2) Now we examine the equation: $$2\sqrt3\sin2\theta=\sqrt3$$
$$\sin2\theta=\frac{\sqrt3}{2\sqrt3}=\frac{1}{2}$$
Over interval $[0^\circ,720^\circ)$, there are 4 values whose sine equals $\frac{1}{2}$, which are $\{30^\circ, 150^\circ, 390^\circ,510^\circ\}$
Therefore, $$2\theta=\{30^\circ, 150^\circ, 390^\circ,510^\circ\}$$
It follows that $$\theta=\{15^\circ, 75^\circ, 195^\circ,255^\circ\}$$
This is the solution set of the equation.
