Question

The sum of roots of the equation ${\sin }^{-1}\frac{3x}{5}+{\sin }^{-1}\frac{4x}{5}={\sin }^{-1}x,x\in [-1,1]$ is

Answer 1

We have,
${\sin }^{-1}\frac{3x}{5}+{\sin }^{-1}\frac{4x}{5}={\sin }^{-1}x,x\in [-1,1]$
Using the formula:
$\{{\sin }^{-1}x+{\sin }^{-1}y={\text{sin}}^{-1}\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\}\}$

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

$\Rightarrow 3x\sqrt{25-16x^2}+4x\sqrt{25-9x^2}=25x$
$\Rightarrow x=0\text{or},3\sqrt{25-16x^2}+4\sqrt{25-9x^2}=25$
Now,
$3\sqrt{25-16x^2}+4\sqrt{25-9x^2}=25$
$\Rightarrow 4\sqrt{25-9x^2}=25-3\sqrt{25-16x^2}$
$\Rightarrow 16(25-9x^2)=625+9(25-16x^2)-150\sqrt{25-16x^2}$
$\Rightarrow 150\sqrt{25-16x^2}=450$
$\Rightarrow 25-16x^2=9\Rightarrow x=\pm 1$
Hence, $x=1,0,-1$ are roots of the given equation.
$\therefore$ Sum of roots $=0$​​​​