Answer
$ y $ is continuous on
$$...\cup(-\frac{\pi}{2},0)\cup (0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi) \cup( \pi,\frac{3\pi}{2}) \cup...$$
Work Step by Step
Given $$ y=\csc 2x=\frac{1}{\sin 2x} $$
Since the of the denominator is zero at
$\sin 2 x=0 \Rightarrow 2x=k \pi \Rightarrow x=k \frac{\pi}{2}, \ \ k \in Z $
So, $ y $ is continuous on
$$...\cup(-\frac{\pi}{2},0)\cup (0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi) \cup( \pi,\frac{3\pi}{2}) \cup...$$
