If sin ^{-1}(4 x)-cos ^{-1}(3 x)=frac{pi}{6}, | vedclass

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(C) Given: $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$....$(i)$Let $A=\sin ^{-1}(4 x)$ and $B=\cos ^{-1}(3 x)$.Then $\sin A=4 x$ and $\cos B=3 x$.

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$\frac{\sqrt{3}}{2 \sqrt{13}}$

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(C) Given: $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$....$(i)$Let $A=\sin ^{-1}(4 x)$ and $B=\cos ^{-1}(3 x)$.Then $\sin A=4 x$ and $\cos B=3 x$.We know that $\cos A=\sqrt{1-16 x^2}$ and $\sin B=\sqrt{1-9 x^2}$.From $(i)$, $A-B=\frac{\pi}{6}$. Taking sine on both sides:$\sin(A-B) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, we get:$(4x)(3x) - \sqrt{1-16 x^2} \sqrt{1-9 x^2} = \frac{1}{2}$.$12x^2 - \frac{1}{2} = \sqrt{(1-16 x^2)(1-9 x^2)}$.Squaring both sides:$(12x^2 - \frac{1}{2})^2 = (1-16 x^2)(1-9 x^2)$.$144x^4 - 12x^2 + \frac{1}{4} = 1 - 9x^2 - 16x^2 + 144x^4$.$-12x^2 + \frac{1}{4} = 1 - 25x^2$.$13x^2 = 1 - \frac{1}{4} = \frac{3}{4}$.$x^2 = \frac{3}{52}$.$x = \sqrt{\frac{3}{52}} = \frac{\sqrt{3}}{2 \sqrt{13}}$.

If $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$, then $x=$

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