If 3 sin^{-1}(frac{2x}{1+x^2}) - 4 cos^{-1}(f | vedclass

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Question

(C) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$.Using the standard trigonometric substitutions for $|x| \le 1$:$\sin^{-1}(\frac{2x}{1+x^2}) = 2

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$1/\sqrt{3}$

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(C) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$.Using the standard trigonometric substitutions for $|x| \le 1$:$\sin^{-1}(\frac{2x}{1+x^2}) = 2 	an^{-1} x = 2	heta$$\cos^{-1}(\frac{1-x^2}{1+x^2}) = 2 	an^{-1} x = 2	heta$$	an^{-1}(\frac{2x}{1-x^2}) = 2 	an^{-1} x = 2	heta$Substituting these into the equation:$3(2	heta) - 4(2	heta) + 2(2	heta) = \frac{\pi}{3}$$6	heta - 8	heta + 4	heta = \frac{\pi}{3}$$2	heta = \frac{\pi}{3}$$	heta = \frac{\pi}{6}$Since $x = 	an 	heta$, we have $x = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.

If $3 \sin^{-1}(\frac{2x}{1+x^2}) - 4 \cos^{-1}(\frac{1-x^2}{1+x^2}) + 2 \tan^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{3}$, then find the value of $x$.

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