#### Answer

$(-\infty, -2] \cup [2, +\infty)$

#### Work Step by Step

RECALL:
The range of the function $f(x)=csc(x)$ is $(-\infty, -1] \cup [1, +\infty)$ because the value of the function will never be between in the interval $(-1, 1)$.
In $f(x)=2\csc{(bx+c)}$, the value of $(bx+c)$ does not affect the range of values of the function.
This means that the range of values of $f(x)=2\csc{(bx+c)}$ will never be in the interval $(-2, 2)$.
Therefore, the range of a function of the form $f(x)=2\csc{(bx+c)}$ is $(-\infty, -2] \cup [2, +\infty)$.