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 $\csc$?
Best Answer
$$f(x)=\dfrac{1+\sin x+\cos x}{\sin x\cos x}$$
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 $\sin x+\cos x-1\le0,f(x)\le2-2\sqrt2$