The number of real roots of the equation sin | vedclass

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(B) Given equation: $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 	an ^{-1} xight)ight\}ight]=0$ $\Rightarrow 2 \cos ^{-1}\left\{\cot \left(2 	a

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$2$

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(B) Given equation: $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 	an ^{-1} xight)ight\}ight]=0$ $\Rightarrow 2 \cos ^{-1}\left\{\cot \left(2 	an ^{-1} xight)ight\}=n \pi, n \in \mathbb{Z}$ $\Rightarrow \cos ^{-1}\left\{\cot \left(2 	an ^{-1} xight)ight\}=\frac{n \pi}{2}$ Since the range of $\cos ^{-1} 	heta$ is $[0, \pi]$, we have $\cos ^{-1}\left\{\cot \left(2 	an ^{-1} xight)ight\} \in \{0, \frac{\pi}{2}, \pi\}$. Case $1$: $\cos ^{-1}\left\{\cot \left(2 	an ^{-1} xight)ight\}=0 \Rightarrow \cot \left(2 	an ^{-1} xight)=1 \Rightarrow 2 	an ^{-1} x = \frac{\pi}{4} + m\pi \Rightarrow 	an ^{-1} x = \frac{\pi}{8} + \frac{m\pi}{2}$. For $x \ge 1$, $	an ^{-1} x \in [\frac{\pi}{4}, \frac{\pi}{2})$. The only value is $	an ^{-1} x = \frac{\pi}{4} \Rightarrow x = 1$. Case $2$: $\cos ^{-1}\left\{\cot \left(2 	an ^{-1} xight)ight\}=\frac{\pi}{2} \Rightarrow \cot \left(2 	an ^{-1} xight)=0 \Rightarrow 2 	an ^{-1} x = \frac{\pi}{2} + m\pi \Rightarrow 	an ^{-1} x = \frac{\pi}{4} + \frac{m\pi}{2}$. For $x \ge 1$, $	an ^{-1} x = \frac{\pi}{4} \Rightarrow x = 1$ (already counted). Case $3$: $\cos ^{-1}\left\{\cot \left(2 	an ^{-1} xight)ight\}=\pi \Rightarrow \cot \left(2 	an ^{-1} xight)=-1 \Rightarrow 2 	an ^{-1} x = \frac{3\pi}{4} + m\pi \Rightarrow 	an ^{-1} x = \frac{3\pi}{8} + \frac{m\pi}{2}$. For $x \ge 1$, $	an ^{-1} x = \frac{3\pi}{8} \Rightarrow x = 	an \frac{3\pi}{8} = \sqrt{2} + 1$. Thus, the roots $\ge 1$ are $1$ and $\sqrt{2} + 1$. The number of such roots is $2$.

The number of real roots of the equation $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$ that are greater than or equal to $1$ are

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