Yes, that's what happens.
$\displaystyle \sin 2x = \frac{\sqrt{3}}{2}}$
Let $\displaystyle 2x=u$ then we obtain the equation $\displaystyle \sin(u) = \frac{\sqrt{3}}{2} \Leftrightarrow \sin u = \sin\left(\frac{\pi}{3}\right)$
We have two different solutions now:
$\displaystyle u = \frac{\pi}{3}+2k\pi \Rightarrow 2x = \frac{\pi}{3}+2k\pi$
$\displaystyle u = \pi - \frac{\pi}{3}+2k\pi \Rightarrow 2x = \frac{2\pi}{3}+2k\pi$
which gives us (if we divide both sides by 2)
$\displaystyle x = \frac{\pi}{6}+k\pi$
$\displaystyle x= \frac{\pi}{3}+k \pi$
where $\displaystyle k \in \mathbb{Z}$