Okay, here we have this:
Considering the provided equation, we are going to find the solution in the interval, so we obtain the following:
[tex]\begin{gathered} \sin \mleft(2\theta\mright)=\frac{\sqrt{3}}{2},\: 0\le\: \theta<\: 2\pi \\ 2\theta=\frac{\pi}{3}+2\pi n,\: 2\theta=\frac{2\pi}{3}+2\pi n \\ \theta=\frac{\pi}{6}+\pi n,\: \theta=\frac{\pi}{3}+\pi n \end{gathered}[/tex]Now we analyze which are the solutions that are within the range, then finally we have:
[tex]\theta=\frac{\pi}{6},\: \theta=\frac{\pi}{3},\: \theta=\frac{7\pi}{6},\: \theta=\frac{4\pi}{3}[/tex]The above values are the solution set of the equation.