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:

Taking sine on both sides $\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^2}{25}}=x\Rightarrow 3x\sqrt{25-16x^2}=25x-4x\sqrt{25-9x^2}$

$\Rightarrow x=0$ or $3\sqrt{25-16x^2}=25-4\sqrt{25-9x^2}$

$\Rightarrow 9(25-16x^2)=625-200\sqrt{25-9x^2}+16(25-9x^2)\Rightarrow 200\sqrt{25-9x^2}=800\Rightarrow \sqrt{25-9x^2}=4$

$\Rightarrow x^2=1\Rightarrow x=\pm 1 andx=0$ 

$\therefore$Total number of solutions = 3.