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(C) Given equation: $\sin ^{-1} \frac{3 x}{5}+\sin ^{-1} \frac{4 x}{5}=\sin ^{-1} x$Using the formula $\sin ^{-1} A + \sin ^{-1} B = \sin ^{-1} (A \sq

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$3$

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(C) Given equation: $\sin ^{-1} \frac{3 x}{5}+\sin ^{-1} \frac{4 x}{5}=\sin ^{-1} x$Using the formula $\sin ^{-1} A + \sin ^{-1} B = \sin ^{-1} (A \sqrt{1-B^2} + B \sqrt{1-A^2})$,we get:$\sin ^{-1}\left(\frac{3 x}{5} \sqrt{1-\frac{16 x^{2}}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^{2}}{25}}ight)=\sin ^{-1} x$Equating the arguments:$\frac{3 x}{5} \sqrt{1-\frac{16 x^{2}}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^{2}}{25}}=x$Case $1$: $x=0$ is a solution.Case $2$: $x 
eq 0$,dividing by $x$:$\frac{3}{5} \sqrt{\frac{25-16 x^{2}}{25}} + \frac{4}{5} \sqrt{\frac{25-9 x^{2}}{25}} = 1$$3 \sqrt{25-16 x^{2}} + 4 \sqrt{25-9 x^{2}} = 25$$4 \sqrt{25-9 x^{2}} = 25 - 3 \sqrt{25-16 x^{2}}$Squaring both sides:$16(25-9 x^{2}) = 625 + 9(25-16 x^{2}) - 150 \sqrt{25-16 x^{2}}$$400 - 144 x^{2} = 625 + 225 - 144 x^{2} - 150 \sqrt{25-16 x^{2}}$$150 \sqrt{25-16 x^{2}} = 450$$\sqrt{25-16 x^{2}} = 3$$25 - 16 x^{2} = 9 \Rightarrow 16 x^{2} = 16 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$Checking $x=1, -1, 0$ in the original equation,all satisfy it.Thus,the number of real values of $x$ is $3$.

Given that the inverse trigonometric functions take principal values only. Then,the number of real values of $x$ which satisfy $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1} x$ is equal to:

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