Answer
$\displaystyle \{\frac{\pi}{3}+2k\pi, \frac{2\pi}{3}+2k\pi, $ (k any integer)$\}$
Work Step by Step
3$\csc x=2\sqrt{3}\qquad/\div 3$
$\displaystyle \csc x=\frac{2\sqrt{3}}{3}$ so
$\displaystyle \sin x=\frac{3}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{3}}{2(3)}=\frac{\sqrt{3}}{2}$
From the unit circle,
$\displaystyle \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$ and $\displaystyle \sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}$
so, since sine has a period of $2\pi,$
the solutions will have form
$\displaystyle \frac{\pi}{3}+2k\pi, $ k any integer
or
$\displaystyle \frac{2\pi}{3}+2k\pi, $ k any integer
Solution set:
$\displaystyle \{\frac{\pi}{3}+2k\pi, \frac{2\pi}{3}+2k\pi, $ (k any integer)$\}$
