Question

The number of non-integral values of $x$ satisfying $\sin \left[2{\cos }^{-1}\right\{\cot \left(2{\tan }^{-1}x\right)\left\}\right]=0$ is

Answer 1

$\sin \left[2{\cos }^{-1}\right\{\cot \left(2{\tan }^{-1}x\right)\left\}\right]=0\Rightarrow 2{\cos }^{-1}\left\{\cot \right(2{\tan }^{-1}x\left)\right\}=n\pi ,n\in Z\Rightarrow \cot \left(2{\tan }^{-1}x\right)=\cos \frac{n\pi }{2}$
We know that,
$\cos \frac{n\pi }{2}=\pm 1\text{or}0$

Case $1$:
$\cot \left(2{\tan }^{-1}x\right)=\pm 1$
We know that,
${\tan }^{-1}x\in (-\frac{\pi }{2},\frac{\pi }{2})\Rightarrow 2{\tan }^{-1}x\in (-\pi ,\pi )$
Now,
$\cot \left(2{\tan }^{-1}x\right)=\pm 1$
$\Rightarrow 2{\tan }^{-1}x=\frac{\pi }{4},\frac{3\pi }{4},-\frac{\pi }{4},-\frac{3\pi }{4}\Rightarrow x=\pm \tan \frac{\pi }{8},\pm \tan \frac{3\pi }{8}$

Case $2$:
$\cot \left(2{\tan }^{-1}x\right)=0\Rightarrow 2{\tan }^{-1}x=\pm \frac{\pi }{2}\Rightarrow x=\pm \tan \left(\frac{\pi }{4}\right)\Rightarrow x=\pm 1$

Hence, four non-integral values of $x$ are possible.