Answer
$y=\csc2x$ is continuous for all $x\ne\frac{k\pi}{2}(k\in Z)$
Work Step by Step
$$y=\csc2x=\frac{1}{\sin2x}$$
- Domain: $y$ is defined where $\sin2x\ne0$, which means $$2x\ne k\pi$$ $$x\ne\frac{k\pi}{2}(k\in Z)$$
- As $x$ approaches any values of $\frac{k\pi}{2}$, $\sin2x$ would approach $0$, thus $\frac{1}{\sin2x}$ will approach infinity, and it does not reach for any definite value.
In other words, $\lim_{x\to(k\pi/2)}\frac{1}{\sin2x}$ does not exist, so the function is discontinuous at all points $x=\frac{k\pi}{2}$.
So for all $x\ne\frac{k\pi}{2}$: $$\lim_{x\to c}\frac{1}{\sin2x}=\frac{1}{\sin2c}$$ Therefore, $y=\csc2x=\frac{1}{\sin2x}$ is continuous in the domain defined.
In conclusion, $y=\csc2x$ is continuous for all $x\ne\frac{k\pi}{2}(k\in Z)$
