For how many distinct values of x,the equatio | vedclass

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(C) Given equation is $\sin [2 \cos^{-1} \cot (2 	an^{-1} x)] = 0$. Let $	heta = 2 	an^{-1} x$. Then $\cot 	heta = \cot (2 	an^{-1} x) = \frac{1

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$6$

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(C) Given equation is $\sin [2 \cos^{-1} \cot (2 	an^{-1} x)] = 0$. Let $	heta = 2 	an^{-1} x$. Then $\cot 	heta = \cot (2 	an^{-1} x) = \frac{1 - 	an^2(	an^{-1} x)}{2 	an(	an^{-1} x)} = \frac{1 - x^2}{2x}$. The equation becomes $\sin [2 \cos^{-1} (\frac{1 - x^2}{2x})] = 0$. This implies $2 \cos^{-1} (\frac{1 - x^2}{2x}) = n\pi$ for some integer $n$. So,$\cos^{-1} (\frac{1 - x^2}{2x}) = \frac{n\pi}{2}$. Taking cosine on both sides,$\frac{1 - x^2}{2x} = \cos(\frac{n\pi}{2})$. For the expression to be defined,the argument of $\cos^{-1}$ must be in $[-1, 1]$,so $|\frac{1 - x^2}{2x}| \le 1$. Possible values for $\cos(\frac{n\pi}{2})$ are $0, 1, -1$. Case $1$: $\frac{1 - x^2}{2x} = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$. Case $2$: $\frac{1 - x^2}{2x} = 1 \Rightarrow x^2 + 2x - 1 = 0 \Rightarrow x = -1 \pm \sqrt{2}$. Case $3$: $\frac{1 - x^2}{2x} = -1 \Rightarrow x^2 - 2x - 1 = 0 \Rightarrow x = 1 \pm \sqrt{2}$. All these $6$ values satisfy the condition $|\frac{1 - x^2}{2x}| \le 1$. Thus,there are $6$ distinct values of $x$.

For how many distinct values of $x$,the equation $\sin [2 \cos^{-1} \cot (2 \tan^{-1} x)] = 0$ holds?

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