The solution set of the equation sin^{-1} x = | vedclass

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(C) Given the equation: $\sin^{-1} x = 2	an^{-1} x$We know that $2	an^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} ight)$ for $|x| \le 1$.Substituti

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$\{-1, 1, 0\}$

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(C) Given the equation: $\sin^{-1} x = 2	an^{-1} x$We know that $2	an^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} ight)$ for $|x| \le 1$.Substituting this into the equation,we get:$\sin^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} ight)$Taking the sine of both sides:$x = \frac{2x}{1+x^2}$Rearranging the terms:$x(1+x^2) = 2x$$x + x^3 = 2x$$x^3 - x = 0$$x(x^2 - 1) = 0$$x(x-1)(x+1) = 0$Thus,the solutions are $x = 0, 1, -1$.Checking the domain: For all these values,$|x| \le 1$,so they are valid solutions.The solution set is $\{-1, 0, 1\}$.

The solution set of the equation $\sin^{-1} x = 2\tan^{-1} x$ is

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