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

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(D) Given the equation: $4 \sin ^{-1} x + 6 \cos ^{-1} x = 3 \pi$ We know the identity: $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$ Rewrite the equa

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(D) Given the equation: $4 \sin ^{-1} x + 6 \cos ^{-1} x = 3 \pi$ We know the identity: $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$ Rewrite the equation as: $4(\sin ^{-1} x + \cos ^{-1} x) + 2 \cos ^{-1} x = 3 \pi$ Substitute the identity: $4(\frac{\pi}{2}) + 2 \cos ^{-1} x = 3 \pi$ Simplify: $2 \pi + 2 \cos ^{-1} x = 3 \pi$ Subtract $2 \pi$ from both sides: $2 \cos ^{-1} x = \pi$ Divide by $2$: $\cos ^{-1} x = \frac{\pi}{2}$ Taking cosine on both sides: $x = \cos(\frac{\pi}{2})$ Therefore: $x = 0$

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

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