Answer
$x=\frac{\pi}{3}+n\pi$ and $x=\frac{2\pi}{3}+n\pi$, where $n$ is an integer.
Work Step by Step
$3~csc^2x=4$
$\frac{3}{4}=\frac{1}{csc^2x}$
$\frac{3}{4}=sin^2x$
$sin~x=±\frac{\sqrt 3}{2}$
The period of $sin~x$ is $2\pi$. The solutions in the interval: $[0,2π)$ are:
$x=\frac{\pi}{3}$, $x=\frac{2\pi}{3}$, $x=\frac{4\pi}{3}$ and $x=\frac{5\pi}{3}$
But, $\frac{4\pi}{3}=\frac{\pi}{3}+\pi$ and $\frac{5\pi}{3}=\frac{2\pi}{3}+\pi$
Now, add multiples of $2\pi$ to each of the solutions:
$x=\frac{\pi}{3}+n\pi$ and $x=\frac{2\pi}{3}+n\pi$, where $n$ is an integer.
