Answer
$x=arcsin\frac{1}{5}+2n\pi$, where $n$ is an integer.
Work Step by Step
$csc^2x-5~csc~x=0$
$csc~x(csc~x-5)=0$
$csc~x=0$. But, there is no $x$ such that $csc~x=0$.
$csc~x-5=0$
$csc~x=5$
$\frac{1}{sin~x}=5$
$sin~x=\frac{1}{5}$
$x=arcsin\frac{1}{5}$
The period of $sin~x$ is $2\pi$. So, add multiples of $2\pi$ to each solution to find the general solution:
$x=arcsin\frac{1}{5}+2n\pi$, where $n$ is an integer.
