Solve 3sin⁻¹(2x/(1+x²)) − 4cos⁻¹((1−x²)/(1+x²)) + 2tan⁻¹(2x/(1−x²))

Solve \( 3\sin^{-1}\left(\frac{2x}{1+x^2}\right) – 4\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + 2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3} \)

Solution:

Use standard identities:

\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}(x) \]

\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x) \]

\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x) \]

Substitute into equation:

\[ 3(2\tan^{-1}x) – 4(2\tan^{-1}x) + 2(2\tan^{-1}x) = \frac{\pi}{3} \]

\[ 6\tan^{-1}x – 8\tan^{-1}x + 4\tan^{-1}x = \frac{\pi}{3} \]

\[ 2\tan^{-1}x = \frac{\pi}{3} \]

\[ \tan^{-1}x = \frac{\pi}{6} \Rightarrow x = \tan\left(\frac{\pi}{6}\right) \]

\[ x = \frac{1}{\sqrt{3}} \]

\[ x = \frac{1}{\sqrt{3}} \]

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