Question

The number of real solutions of the equation
${\sin }^{-1}\left(\sum _{i=1}^{\infty }{x}^{i+1}-x\sum _{i=1}^{\infty }{\left(\frac{x}{2}\right)}^i\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\sum _{i=1}^{\infty }{(-\frac{x}{2})}^i-\sum _{i=1}^{\infty }(-x{)}^i\right)$ lying in the interval $(-\frac{1}{2},\frac{1}{2})$ is

Answer 1

$\sum _{i=1}^{\infty }{x}^{i+1}=\frac{x^2}{1-x}$
$\sum _{i=1}^{\infty }{\left(\frac{x}{2}\right)}^i=\frac{x}{2-x}$
$\sum _{i=1}^{\infty }{(-\frac{x}{2})}^i=\frac{-x}{2+x}$
$\sum _{i=1}^{\infty }(-x{)}^i=\frac{-x}{1+x}$
To have real solutions
$\sum _{i=1}^{\infty }{x}^{i+1}-x\sum _{i=1}^{\infty }{\left(\frac{x}{2}\right)}^i=\sum _{i=1}^{\infty }{\left(\frac{-x}{2}\right)}^i-\sum _{i=1}^{\infty }(-x{)}^i$
$\frac{x^2}{1-x}-\frac{x^2}{2-x}=\frac{-x}{2+x}+\frac{x}{1+x}$
$x(x^3+2x^2+5x-2)=0$
$\therefore x=0$ and let $f\left(x\right)=x^3+2x^2+5x-2$
$f\left(\frac{1}{2}\right)\cdot f\left(-\frac{1}{2}\right)<0$
Hence two solutions exist.