${\sin ^{ - 1}}{{3x} \over 5} + {\sin ^{ - 1}}{{4x} \over 5} = {\sin ^{ - 1}}x$ ${\sin ^{ - 1}}\left( {{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} } \right) = {\sin ^{ - 1}}x$ ${{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} = x$ $x = 0$ or $3\sqrt {25 - 16{x^2}} + 4\sqrt {25 - 9{x^2}} = 25$ $4\sqrt {25 - 9{x^2}} = 25 - 3\sqrt {25 - 16{x^2}}$ Squaring we get $16(25 - 9{x^2}) = 625 - 9(25 - 16{x^2}) - 150\sqrt {25 - 16{x^2}}$ $400 = 625 + 225 - 150\sqrt {25 - 16{x^2}}$ $\sqrt {25 - 16{x^2}} = 3 \Rightarrow 25 - 16{x^2} = 9$ $\Rightarrow {x^2} = 1$ Put x = 0, 1, $-$1 in the original equation We see that all values satisfy the original equation. Number of solution = 3