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 $\csc$?

First of all, we need
$\csc2x$ must be finite $\implies\sin2x\ne0\implies2x\ne n\pi\  \ \ \  (1)$ where $n$ is any integer
Now if $\sin x+\cos x+1=0,$
using double angle formula, $$\cos\dfrac x2\left(\sin\dfrac x2+\cos\dfrac x2\right)=0$$
$\cos\dfrac x2=0\iff x=(2r+1)\pi\  \ \ \  (2)$
and $\sin\dfrac x2+\cos\dfrac x2=0\implies x=2r\pi-\dfrac\pi2\  \ \ \  (3)$
Both $(2),(3)$ are been precluded by $(1)$
Let $\sin x+\cos x+1=k\implies k\ne0$
$\sin x\cos x=\dfrac{(\sin x+\cos x)^2-1}2=\dfrac{k^2-2k}2$
$f(x)=\dfrac{2k}{k(k-2)}=\dfrac2{k-2}$
As $k\ne0, f(x)\ne\dfrac2{0-2}\  \ \ \  (4)$
Now $-\sqrt2\le\sin x+\cos x\le\sqrt2\implies-\sqrt2-1\le k-2\le\sqrt2-1$
If $0< k\le\sqrt2-1, f(x)\ge\dfrac2{\sqrt2-1}=?$
If $0>k\ge-\sqrt2-1,  f(x)\le\dfrac2{-(\sqrt2+1)}=?$
But remember $(4)$
