Finding the range of $f(x)=2\csc(2x)+\sec x+\csc x$ Hi this is the question:

Find the range of $$f(x)=2\csc(2x)+\sec x+\csc x$$

What I've tried:
I know that the range of $\csc(x)$ which is $R\setminus (-1,1) $, the range of $\sec(x)$ is $R\setminus (-1,1)$ too. And I've managed to simplify the expression to have in terms of $\sec x$ and $\csc x$ as such:
$$f(x)=\sec(x)\cdot \csc(x)+\sec(x)+\csc(x).$$
But then, what do I do when I have 3 terms and not just $\sec$ or $\csc$?

Starting with $$f(x)=\frac{2}{\sin x+ \cos x-1} \implies y=\frac{1+t^2}{t-t^2} \implies (1+y)t^2-yt-1=0,$$
As $t=\tan(x/2)$ lies in $(-\infty, \infty)$, the range will be all values of $y$ when the above quadratic has real roots: $B^2 \ge 4AC$
$$\implies y^2-4y-4 \ge 0 \implies y\ge 2+\sqrt{2} ~or~ y\le2-\sqrt{2}$$
