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 $\csc$?
 A: $$f(x)=\dfrac{1+\sin x+\cos x}{\sin x\cos x}$$
Clearly $f(x)$ is undefined for $2x=n\pi$ where $n$ is any integer
$$\dfrac{f(x)}2=\dfrac{1+\sin x+\cos x}{(\sin x+\cos x)^2-1}=\dfrac1{\sin x+\cos x-1} $$
Now use $-\sqrt2\le\sin x+\cos x\le\sqrt2$ to find
$$\implies-\sqrt2-1\le\sin x+\cos x-1\le\sqrt2-1$$
If $\sin x+\cos x-1\ge0, f(x)\ge2+2\sqrt2$
and if $\sin x+\cos x-1\le0,f(x)\le2-2\sqrt2$
A: The derivative of the secant function is $\sin x/\cos^2x$; the derivative of the cosecant function is $-\cos x/\sin^2x$, so you get
$$
f'(x)=-4\frac{\cos2x}{\sin^22x}+\frac{\sin x}{\cos^2x}-\frac{\cos x}{\sin^2x}
=\frac{\sin^2x-\cos^2x+\sin^3x-\cos^3x}{\sin^2x\cos^2x}
$$
We can disregard the denominator and factor the numerator as
$$
(\sin x-\cos x)(\sin x+\cos x+\sin^2x+\sin x\cos x+\cos^2x)
=(\sin x-\cos x)(1+\sin x)(1+\cos x)
$$
Quite nicer! This is positive where $\sin x>\cos x$. Limiting ourselves to the interval $(0,2\pi)$, but also noting that the function is undefined at integer multiples of $\pi2$, we see that this happens for $\pi/4<x<5\pi/4$. Thus our function is

*

*decreasing over $(0,\pi/4)$

*increasing over $(\pi/4,\pi/2)$

*increasing over $(\pi/2,5\pi/4)$

*decreasing over $(5\pi/4,3\pi/2)$

*decreasing over $(3\pi/2,2\pi)$
Thus, taking into account the asymptotes, we see that the branch in $(0,\pi/2)$ has range $(f(\pi/4),\infty)=(2+2\sqrt{2},\infty)$.
The branch in $(\pi/2,2\pi)$ is a bit more complicated, because the function doesn't have asymptotes at $\pi$ and at $3\pi/2$. Indeed,
$$
\lim_{x\to\pi}f(x)=-1=\lim_{x\to3\pi/2}f(x)
$$
If we extend the function by continuity, the range over $(\pi/2,2\pi)$ would be $(-\infty,f(5\pi/4))=(-\infty,2-2\sqrt{2})$.
As the function is not defined at $\pi$ and $3\pi/2$, the range is
$$
(-\infty,-1)\cup(-1,2-2\sqrt{2})\cup(2+2\sqrt{2},\infty)
$$

A: 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: $B^2 \ge 4AC$
$$\implies y^2-4y-4 \ge 0 \implies y\ge 2+\sqrt{2} ~or~ y\le2-\sqrt{2}$$
A: 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 $(4)$
