Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy

$${\sin ^{ - 1}}\left( {{{3x} \over 5}} \right) + {\sin ^{ - 1}}\left( {{{4x} \over 5}} \right) = {\sin ^{ - 1}}x$$ is equal to :

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