If 2 sin ^{-1} x-3 cos ^{-1} x=4, x in[-1,1], | vedclass

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(B) Given that $2 \sin ^{-1} x - 3 \cos ^{-1} x = 4$. We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,so $\cos ^{-1} x = \frac{\pi}{2} - \s

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$\frac{6 \pi-4}{5}$

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(B) Given that $2 \sin ^{-1} x - 3 \cos ^{-1} x = 4$. We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,so $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$. Substituting this into the given equation: $2 \sin ^{-1} x - 3(\frac{\pi}{2} - \sin ^{-1} x) = 4$ $2 \sin ^{-1} x - \frac{3 \pi}{2} + 3 \sin ^{-1} x = 4$ $5 \sin ^{-1} x = 4 + \frac{3 \pi}{2}$ $\sin ^{-1} x = \frac{8 + 3 \pi}{10}$ Now,we need to find the value of $2 \sin ^{-1} x + 3 \cos ^{-1} x$. $2 \sin ^{-1} x + 3 \cos ^{-1} x = 2 \sin ^{-1} x + 3(\frac{\pi}{2} - \sin ^{-1} x)$ $= 2 \sin ^{-1} x + \frac{3 \pi}{2} - 3 \sin ^{-1} x$ $= \frac{3 \pi}{2} - \sin ^{-1} x$ $= \frac{3 \pi}{2} - \frac{8 + 3 \pi}{10}$ $= \frac{15 \pi - 8 - 3 \pi}{10}$ $= \frac{12 \pi - 8}{10} = \frac{6 \pi - 4}{5}$

If $2 \sin ^{-1} x-3 \cos ^{-1} x=4, x \in[-1,1]$,then $2 \sin ^{-1} x+3 \cos ^{-1} x$ is equal to

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