For x in (-1, 1],the number of solutions of t | vedclass

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(B) Given the equation $\sin^{-1} x = 2 	an^{-1} x$.Let $	an^{-1} x = 	heta$,then $x = 	an 	heta$,where $	heta \in (-\pi/4, \pi/4]$ since $x \in

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$2$

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(B) Given the equation $\sin^{-1} x = 2 	an^{-1} x$.Let $	an^{-1} x = 	heta$,then $x = 	an 	heta$,where $	heta \in (-\pi/4, \pi/4]$ since $x \in (-1, 1]$.The equation becomes $\sin^{-1}(	an 	heta) = 2	heta$.Taking $\sin$ on both sides,we get $	an 	heta = \sin(2	heta)$.$	an 	heta = 2 \sin 	heta \cos 	heta$.$\frac{\sin 	heta}{\cos 	heta} = 2 \sin 	heta \cos 	heta$.$\sin 	heta (\frac{1}{\cos 	heta} - 2 \cos 	heta) = 0$.This gives $\sin 	heta = 0$ or $\frac{1}{\cos 	heta} = 2 \cos 	heta$.Case $1$: $\sin 	heta = 0 \implies 	heta = 0$,which gives $x = 	an 0 = 0$.Case $2$: $2 \cos^2 	heta = 1 \implies \cos^2 	heta = 1/2 \implies \cos 	heta = \pm 1/\sqrt{2}$.Since $	heta \in (-\pi/4, \pi/4]$,$\cos 	heta$ must be positive,so $\cos 	heta = 1/\sqrt{2}$.This implies $	heta = \pi/4$ or $	heta = -\pi/4$.If $	heta = \pi/4$,$x = 	an(\pi/4) = 1$.If $	heta = -\pi/4$,$x = 	an(-\pi/4) = -1$,but the domain is $x \in (-1, 1]$,so $x = -1$ is excluded.Thus,the solutions are $x = 0$ and $x = 1$. The number of solutions is $2$.

For $x \in (-1, 1]$,the number of solutions of the equation $\sin^{-1} x = 2 \tan^{-1} x$ is equal to

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