How to handle all cases imposed on $x$ so as to solve $\sin({2\cos^{-1}({\cot({2\tan^{-1}x})})}) = 0$?

Question

Given equation to solve: $$\sin({2\cos^{-1}({\cot({2\tan^{-1}x})})}) = 0$$

What I considered was as $2\arctan x = \arctan(\frac{2x}{1-x^2})$ only when $|x|\leq 1$. So now, after simplifying that, we reach $2\arccos(...)$, where again we have to use a condition on $x$, so in general there are four ($2\times 2$) situations where $x$ can lie and restricted, and we have to check each case and also terminate the extraneuous roots whenever possible.

My book just considered a simple case where there is nothing $-\pi$ or $\pi$m and didn't even bother to impose conditions like $|x|\leq 1$ (even though they were considering just that case only).

Why is it possible to just consider that unique case and that will allow us to solve this equation even without bothering about $x$'s nature?

I have posted the solution too given so that if it's totally wrong it should be corrected. As nothing about the nature of $x$ is considered.

Answer 1

You're right that $$2\arctan x = \arctan\left(\frac{2x}{1-x^2}\right)\implies|x|<1;$$ note also that if the principal range of $\operatorname{arccot}$ is taken to be $(0,\pi),$ then $\operatorname{arccot}(-7)\ne\arctan\left(\frac{1}{-7}\right).$

Noting that $$\tan\left(2\arctan x\right) \equiv\dfrac{2x}{1-x^2}$$ and that $\arccos$ has principal range $[0,\pi]$ and domain $[-1,1],$ here's a clearer answer:

\begin{align}&\sin\left(2\arccos\left(\cot\left(2\arctan\left(x\right)\right)\right)\right)=0\\ \iff&\sin\left(2\arccos\left( \frac{1-x^2}{2x} \right)\right)=0\\ \iff&\arccos\left( \frac{1-x^2}{2x} \right)=0,\frac\pi2,\pi\\ \iff& \frac{1-x^2}{2x} =1,0,-1\quad\text{and}\quad \frac{1-x^2}{2x}\in[-1,1]\\ \iff& \frac{1-x^2}{2x} =1,0,-1\\ \iff&\cdots\\ \iff& x =-1\pm\sqrt2,\,\pm1,\,1\pm\sqrt2.\end{align}