Given that the inverse trigonometric functions take principal...

$\sin^{- 1}\dfrac{3x}{5} + \sin^{- 1}\dfrac{4x}{5} = \sin^{- 1}x$

$\sin^{- 1}\left(\dfrac{3x}{5}\sqrt{1 - \dfrac{16{x}^{2}}{25}} + \dfrac{4x}{5}\sqrt{1 - \dfrac{9{x}^{2}}{25}}\right) = \sin^{- 1}x$

$\dfrac{3x}{5}\sqrt{1 - \dfrac{16{x}^{2}}{25}} + \dfrac{4x}{5}\sqrt{1 - \dfrac{9{x}^{2}}{25}} = x$

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

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

squaring we get

$16\left(25 - 9{x}^{2}\right) = 625 + 9\left(25 - 16{x}^{2}\right) - 150\sqrt{25 - 16{x}^{2}}$

$400 = 625 + 225 - 150\sqrt{25 - 16{x}^{2}}$

$\sqrt{25 - 16{x}^{2}} = 3 ⇒ 25 - 16{x}^{2} = 9$

$⇒ {x}^{2} = 1$

Put $x = 0, 1, - 1$ in the original equation We see that all values satisfy the original equation.

Number of solution $= 3$