Answer
The solution set is
$$\{\frac{\pi}{3}+2\pi n, \frac{2\pi}{3}+2\pi n, n\in Z\}$$
Work Step by Step
$$3\csc x-2\sqrt3=0$$
1) Solve the equation over the interval $[0,2\pi)$
$$3\csc x-2\sqrt3=0$$
$$\csc x=\frac{2\sqrt3}{3}$$
Since $\sin x=\frac{1}{\csc x}$, this also equals to $$\sin x=\frac{3}{2\sqrt3}=\frac{\sqrt3}{2}$$
Over the interval $[0,2\pi)$, there are two values of $x$ where $\csc x=\frac{2\sqrt3}{3}$ or equally $\sin x=\frac{\sqrt3}{2}$, which are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$.
2) Solve the equation for all solutions
- The integer multiples of the period of the cosecant function is $2\pi$.
- We apply it to each solution found in step 1. The results are the solution set.
So the solution set is
$$\{\frac{\pi}{3}+2\pi n, \frac{2\pi}{3}+2\pi n, n\in Z\}$$
