The sum of the values of x satisfying the equ | vedclass

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(A) Given equation: $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1}(x)$.Taking $\sin$ on both sides:$\frac{3x}{5

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(A) Given equation: $\sin ^{-1}\left(\frac{3 x}{5}ight)+\sin ^{-1}\left(\frac{4 x}{5}ight)=\sin ^{-1}(x)$.Taking $\sin$ on both sides:$\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}} + \frac{4x}{5}\sqrt{1-\frac{9x^2}{25}} = x$.If $x=0$,the equation holds true.For $x 
eq 0$,divide by $x$:$\frac{3}{25}\sqrt{25-16x^2} + \frac{4}{25}\sqrt{25-9x^2} = 1$.$3\sqrt{25-16x^2} + 4\sqrt{25-9x^2} = 25$.Let $3\sqrt{25-16x^2} = 25 - 4\sqrt{25-9x^2}$.Squaring both sides:$9(25-16x^2) = 625 + 16(25-9x^2) - 200\sqrt{25-9x^2}$.$225 - 144x^2 = 625 + 400 - 144x^2 - 200\sqrt{25-9x^2}$.$200\sqrt{25-9x^2} = 800$.$\sqrt{25-9x^2} = 4$.$25-9x^2 = 16 \Rightarrow 9x^2 = 9 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.Checking $x=1$: $\sin^{-1}(3/5) + \sin^{-1}(4/5) = \sin^{-1}(1) = \pi/2$. This is true.Checking $x=-1$: $\sin^{-1}(-3/5) + \sin^{-1}(-4/5) = -(\sin^{-1}(3/5) + \sin^{-1}(4/5)) = -\pi/2 = \sin^{-1}(-1)$. This is true.The values of $x$ are $0, 1, -1$. The sum is $0 + 1 + (-1) = 0$.

The sum of the values of $x$ satisfying the equation $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1}(x)$ is:

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