Let the inverse trigonometric functions take | vedclass

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(A) We are given the equation $2 \sin ^{-1} x + 3 \cos ^{-1} x = \frac{2 \pi}{5}$.We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$, which im

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(A) We are given the equation $2 \sin ^{-1} x + 3 \cos ^{-1} x = \frac{2 \pi}{5}$.We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$, which implies $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$.Substituting this into the given equation:$2(\frac{\pi}{2} - \cos ^{-1} x) + 3 \cos ^{-1} x = \frac{2 \pi}{5}$$\pi - 2 \cos ^{-1} x + 3 \cos ^{-1} x = \frac{2 \pi}{5}$$\pi + \cos ^{-1} x = \frac{2 \pi}{5}$$\cos ^{-1} x = \frac{2 \pi}{5} - \pi$$\cos ^{-1} x = -\frac{3 \pi}{5}$Since the range of the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$, the value $-\frac{3 \pi}{5}$ is outside this range.Therefore, there is no real value of $x$ that satisfies the equation.The number of real solutions is $0$.

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin ^{-1} x + 3 \cos ^{-1} x = \frac{2 \pi}{5}$ is:

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