Given equation
$\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$
$\Rightarrow 2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=n \pi, n \in \mathbf{Z}$
$\Rightarrow \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=\frac{n \pi}{2}$
$\because \quad \cos ^{-1} x \in[0, \pi]$
$\therefore \quad \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=0, \frac{\pi}{2}, \pi$
$\Rightarrow \cot \left(2 \tan ^{-1} x\right)=1,0,-1 \Rightarrow 2 \tan ^{-1} x=m \pi+\frac{\pi}{4}$
$m \pi+\frac{\pi}{2}$ or $m \pi-\frac{\pi}{4}, m \in \mathbf{Z}$
$\Rightarrow \tan ^{-1} x=\frac{m \pi}{2} \pm \frac{\pi}{8}$ or $\frac{m \pi}{2}+\frac{\pi}{4}, m \in \mathbf{Z}$
$\because \quad \tan ^{-1} x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\therefore \tan ^{-1} x= \pm \frac{\pi}{8}, \frac{\pi}{2}-\frac{\pi}{8},-\frac{\pi}{2}+\frac{\pi}{8}, \pm \frac{\pi}{4}$
$\therefore \quad x= \pm \tan \frac{\pi}{8}, \pm \cot \frac{\pi}{8}, \pm \tan \frac{\pi}{4}$
$\Rightarrow \quad x= \pm(\sqrt{2}-1), \pm \frac{1}{\sqrt{2}-1}, \pm 1$
$= \pm(\sqrt{2}-1), \pm(\sqrt{2}+1), \pm 1$
The roots which are greater than or equal to one are $\sqrt{2}+1,1$.