Given that the inverse trigonometric function take principal value only. Then the number   

          of real value of x which satisfy  ${\sin }^{-1}\left(\frac{3x}{5}\right)+{\sin }^{-1}\left(\frac{4x}{5}\right)={\sin }^{-1}x$is equal to:

$\begin{array}{l}\text{Using sum of two inverse of sine functions and then remove sine from both sides, we get }\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^2}{25}}=x\\ \Rightarrow 3x\sqrt{25-16x^2}=25x-4x\sqrt{25-9x^2}\\ \Rightarrow x=0\text{ or }\\ \Rightarrow 3\sqrt{25-9x^2}=25-4\sqrt{25-9x^2}\\ squaring on both sides\\ \Rightarrow 9\left(25-9x^2\right)=625-200\sqrt{25-9x^2}+16\left(25-9x^2\right)\\ Simplifying the above equation we get 200\sqrt{25-9x^2}=800\\ \Rightarrow x^2=1 \Rightarrow x=\pm 1\text{ and }x=0 \therefore \text{ Total number of solutions }=3\end{array}$