[tex] \bf \cfrac{\theta }{2}=sin^{-1}\left( \frac{1}{2} \right)\qquad \textit{let's say that }sin^{-1}\left( \frac{1}{2} \right)=\alpha\\\\\\\cfrac{\theta }{2}=\alpha\implies \theta =2\alpha\\\\------------------------------- [/tex]
[tex] \bf sin^{-1}\left( \frac{1}{2} \right)=\alpha\implies sin(\alpha)=\cfrac{1}{2}\implies \alpha=\begin{cases}\frac{\pi }{6}\\\\\frac{5\pi }{6}\end{cases}\\\\\\\theta =2\alpha\implies \theta =\begin{cases}2\cdot \frac{\pi }{6}\implies \frac{\pi }{3}\\\\2\cdot \frac{5\pi }{6}\implies \frac{5\pi }{3}\end{cases}\\\\\\\stackrel{\textit{all solutions for }\theta }{\pm\cfrac{\pi }{3}\pm\cfrac{4\pi n}{3}}\qquad \{n~|~ n\in \mathbb{Z}\} [/tex]