$\sin x = \dfrac{\sqrt 3}2$
$⇔ \sin x = \sin \dfrac{\pi}3$
\(⇔ \left[ \begin{array}{l}x=\dfrac{\pi}3+k2\pi\\x=\dfrac{2\pi}3+k2\pi\end{array} \right. (k\in \mathbb{Z})\)
Vậy $S=\left\{\dfrac{\pi}3+k2\pi ; \dfrac{2\pi}3+k2\pi \; \Bigg | \; k\in \mathbb{Z}\right\}$