If 2 tan^{-1} x = 3 sin^{-1} x and x neq 0, t | vedclass

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(D) Given $2 	an^{-1} x = 3 \sin^{-1} x$. Using the identity $2 	an^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} ight)$ and $3 \sin^{-1} x = \sin^{-

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$\sqrt{17}$

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(D) Given $2 	an^{-1} x = 3 \sin^{-1} x$. Using the identity $2 	an^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} ight)$ and $3 \sin^{-1} x = \sin^{-1} (3x - 4x^3)$, we have: $\sin^{-1} \left( \frac{2x}{1+x^2} ight) = \sin^{-1} (3x - 4x^3)$ $\Rightarrow \frac{2x}{1+x^2} = 3x - 4x^3$ Since $x 
eq 0$, we can divide by $x$: $\frac{2}{1+x^2} = 3 - 4x^2$ $2 = (3 - 4x^2)(1 + x^2)$ $2 = 3 + 3x^2 - 4x^2 - 4x^4$ $4x^4 + x^2 - 1 = 0$ Using the quadratic formula for $x^2$: $x^2 = \frac{-1 \pm \sqrt{1^2 - 4(4)(-1)}}{2(4)} = \frac{-1 \pm \sqrt{17}}{8}$ Since $x^2 > 0$, we take $x^2 = \frac{\sqrt{17} - 1}{8}$. Then $8x^2 = \sqrt{17} - 1$, which implies $8x^2 + 1 = \sqrt{17}$.

If $2 \tan^{-1} x = 3 \sin^{-1} x$ and $x \neq 0$, then $8x^2 + 1 =$

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