For how many distinct values of x , the equation sin 2 cos - 1 cot 2 tan - 1 x = 0 holds?

Given, $\sin \left[2{\cos }^{-1}\cot \left(2{\tan }^{-1}x\right)\right]=0$

Let ${\tan }^{-1}x=y\Rightarrow x=\tan y$

Then, $\sin \left[2{\cos }^{-1}\cot \left(2y\right)\right]=0$

$\Rightarrow \sin \left[2{\cos }^{-1}\left(\frac{1-{\tan }^{2}y}{2\tan y}\right)\right]=0$

$\Rightarrow \sin \left[2{\cos }^{-1}\left(\frac{1-x^2}{2x}\right)\right]=0$

$\Rightarrow \sin \left[{\cos }^{-1}\left(2{\left(\frac{1-x^2}{2x}\right)}^2-1\right)\right]=0$

$\Rightarrow \sin \left[{\sin }^{-1}\sqrt{1-{\left(2{\left(\frac{1-x^2}{2x}\right)}^2-1\right)}^2}\right]=0$

$\Rightarrow \sqrt{1-{\left(2{\left(\frac{1-x^2}{2x}\right)}^2-1\right)}^2}=0$

$\Rightarrow 1-{\left(2{\left(\frac{1-x^2}{2x}\right)}^2-1\right)}^2=0$

$\Rightarrow {\left(2{\left(\frac{1-x^2}{2x}\right)}^2-1\right)}^2=1$

$\Rightarrow \left(2{\left(\frac{1-x^2}{2x}\right)}^2-1\right)=\pm 1$

$\Rightarrow 2{\left(\frac{1-x^2}{2x}\right)}^2=1\pm 1=2, 0$

$\Rightarrow 2{\left(\frac{1-x^2}{2x}\right)}^2=2$ or $2{\left(\frac{1-x^2}{2x}\right)}^2=0$

$\Rightarrow {\left(\frac{1-x^2}{2x}\right)}^2=1$ or $1-x^2=0$

$\Rightarrow \left(\frac{1-x^2}{2x}\right)=\pm 1$ or $x=\pm 1$

$\Rightarrow x^2+2x-1=0, x^2-2x-1=0$ or $x=\pm 1$

$\Rightarrow x=-1\pm \sqrt{2}, x=1\pm \sqrt{2}$ or $x=\pm 1$

Hence, there are $6$possible values of $x$.