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 $(0,2)$ is _____.
(Here, the inverse trigonometric function ${\sin }^{-1}x$  and  ${\cos }^{-1}x$  assume values in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ and $[0,\pi ]$ , respectively.)
 

${\sin }^{-1}(\frac{x^2}{1-x}-\frac{x\left(\frac{x}{2}\right)}{1-\frac{x}{2}})=\frac{\pi }{2}-{\cos }^{-1}(\frac{-x/2}{1+\frac{x}{2}}-\frac{(-x)}{1+x})$

$\Rightarrow {\sin }^{-1}\left(x^2(\frac{1}{1-x}-\frac{1}{2-x})\right)=\frac{\pi }{2}-{\cos }^{-1}\left(x(\frac{1}{1+x}-\frac{1}{2+x})\right)$

${\sin }^{-1}\left[\frac{x^2}{(1-x)(2-x)}\right]=\frac{\pi }{2}-{\cos }^{-1}\left[\frac{x}{(1+x)(2+x)}\right]$

${\sin }^{-1}\left[\frac{x^2}{(1-x)(2-x)}\right]={\sin }^{-1}\left[\frac{x}{(1+x)(2+x)}\right]$

$\Rightarrow x[\frac{x}{(1-x)(2-x)}-\frac{1}{(1+x)(2+x)}]=0 x=0 or x^3+3x^2+2x=x^2-3x+2$

$\Rightarrow x^3+2x^2+5x-2=0 increa\sin g function \forall x$

$f\left(0\right)=-2<0, f\left(\frac{1}{2}\right)>0$

$one root between (0,\frac{1}{2}) total number of solutions = 1$