## Precalculus (6th Edition)

$\displaystyle \{\frac{\pi}{3}+2k\pi, \frac{2\pi}{3}+2k\pi,$ (k any integer)$\}$
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)$\}$