The number of real solutions of the equation sin - 1 ⁡ ∑ i = 1 ∞ x i + 1 - x ∑ i = 1…

$\sum _{i=1}^{\infty }{x}^{i+1}=\frac{x^2}{1-x}$
$\sum _{i=1}^{\infty }{\left(\frac{x}{2}\right)}^i=\frac{x}{2-x}$
$\sum _{i=1}^{\infty }{\left(-\frac{x}{2}\right)}^i=\frac{-x}{2+x}$
$\sum _{i=1}^{\infty }{\left(-x\right)}^i=\frac{-x}{1+x}$
For the given equation to have real solutions
$\sum _{i=1}^{\infty }{x}^{i+1}-x\sum _{i=1}^{\infty }{\left(\frac{x}{2}\right)}^i=\sum _{i=1}^{\infty }{\left(-\frac{x}{2}\right)}^i-\sum _{i=1}^{\infty }{\left(-x\right)}^i$
$\frac{x^2}{1-x}-\frac{x^2}{2-x}=\frac{-x}{2+x}+\frac{x}{1+x}$
$x\left(x^3+2x^2+5x-2\right)=0$

$\Rightarrow x=0 is a solution$$,also$

$$\begin{gathered} f\left(x\right)=x^3+2x^2+5x-2 \\ \Rightarrow f^{\text{'}}\left(x\right)=3x^2+4x+5 \\ D=16-60=-44<0 \\ \Rightarrow f is strictly increa\sin g \end{gathered}$$
$and \because f\left(\frac{1}{2}\right).f\left(-\frac{1}{2}\right)<0$
$Hence two solution exist.$