Given that the inverse trigonometric functions take principal values only.

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

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

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

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

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

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

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

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

$\Rightarrow \quad x^{2}=1$

Put $\mathrm{x}=0,1,-1$ in the original equation

We see that all values satisfy the original equation.

Number of solution $=3$