Originally Posted by
SlipEternal For what values of $\theta$ do you have $\sin \theta = \dfrac{1}{2}$? You should know the table:
$\begin{matrix}\theta & \sin \theta \\ 0 & 0 \\ \dfrac{\pi}{6} & \dfrac{1}{2} \\ \dfrac{\pi}{4} & \dfrac{\sqrt{2}}{2} \\ \dfrac{\pi}{3} & \dfrac{\sqrt{3}}{2} \\ \dfrac{\pi}{2} & 1\end{matrix}$
Now, in the interval $x \in (-\pi,\pi)$, will you have any other possible values for $\theta$? How can you tell? Once you have all possible values for $\theta$, you have $\theta = \dfrac{3x}{2}$. Solve for $x$. (Hint: you should find three possible values for $\theta$ where $x \in (-\pi,\pi)$).