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(D) Given equation is $\sin^{-1} x = 2 	an^{-1} x$. Let $	an^{-1} x = 	heta$,then $x = 	an 	heta$. Substituting this into the equation,we get $\s

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$3$

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(D) Given equation is $\sin^{-1} x = 2 	an^{-1} x$. Let $	an^{-1} x = 	heta$,then $x = 	an 	heta$. Substituting this into the equation,we get $\sin^{-1}(	an 	heta) = 2	heta$. Taking $\sin$ on both sides,$	an 	heta = \sin(2	heta)$. Using the identity $\sin(2	heta) = \frac{2 	an 	heta}{1 + 	an^2 	heta}$,we have $	an 	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$. This implies $	an 	heta (1 - \frac{2}{1 + 	an^2 	heta}) = 0$. So,$	an 	heta = 0$ or $1 + 	an^2 	heta = 2$,which means $	an^2 	heta = 1$. If $	an 	heta = 0$,then $x = 0$. If $	an^2 	heta = 1$,then $	an 	heta = 1$ or $	an 	heta = -1$,so $x = 1$ or $x = -1$. Checking these values in the original equation: For $x = 0$: $\sin^{-1}(0) = 0$ and $2 	an^{-1}(0) = 0$. (Valid) For $x = 1$: $\sin^{-1}(1) = \frac{\pi}{2}$ and $2 	an^{-1}(1) = 2(\frac{\pi}{4}) = \frac{\pi}{2}$. (Valid) For $x = -1$: $\sin^{-1}(-1) = -\frac{\pi}{2}$ and $2 	an^{-1}(-1) = 2(-\frac{\pi}{4}) = -\frac{\pi}{2}$. (Valid) Thus,there are $3$ solutions: $x \in \{-1, 0, 1\}$.

Find the number of solutions for the equation $\sin^{-1} x = 2 \tan^{-1} x$ (in principal values).

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