If 4 sin ^{-1} x + 6 cos ^{-1} x = 3 pi,then | vedclass

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(C) Given equation: $4 \sin ^{-1} x + 6 \cos ^{-1} x = 3 \pi$. We know the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,which implies $\cos

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(C) Given equation: $4 \sin ^{-1} x + 6 \cos ^{-1} x = 3 \pi$. We know the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,which implies $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$. Substituting this into the equation: $4 \sin ^{-1} x + 6(\frac{\pi}{2} - \sin ^{-1} x) = 3 \pi$ $4 \sin ^{-1} x + 3 \pi - 6 \sin ^{-1} x = 3 \pi$ $-2 \sin ^{-1} x + 3 \pi = 3 \pi$ $-2 \sin ^{-1} x = 0$ $\sin ^{-1} x = 0$ $x = \sin(0) = 0$.

If $4 \sin ^{-1} x + 6 \cos ^{-1} x = 3 \pi$,then $x = \ldots$.

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