Question

The number of real solutions of the equation
${\sin }^{-1}(\sum _{i=1}^{\infty }{x}^{i+1}-x\sum _{i=1}^{\infty }{\left(\frac{x}{2}\right)}^i)=\frac{\pi }{2}-{\cos }^{-1}(\sum _{i=1}^{\infty }{(-\frac{x}{2})}^i-\sum _{i=1}^{\infty }(-x{)}^i)$
lying in the interval $(-\frac{1}{2},\frac{1}{2})$ is .
(Here, the inverse trigonometric functions ${\sin }^{-1}x$ and
${\cos }^{-1}x$ assume values in $[-\frac{\pi }{2},\frac{\pi }{2}]$ and $[0,\pi ]$, respectively.)

Answer 1

$x$ lies in interval $(-\frac{1}{2},\frac{1}{2})$
L.H.S,
${\sin }^{-1}(\sum _{i=1}^{\infty }{x}^{i+1}-x\sum _{i=1}^{\infty }{\left(\frac{x}{2}\right)}^i)={\sin }^{-1}(\frac{x^2}{1-x}-x\cdot \frac{x/2}{1-\left(x/2\right)})={\sin }^{-1}(\frac{x^2}{1-x}-\frac{x^2}{2-x})$
Now R.H.S,
$\frac{\pi }{2}-{\cos }^{-1}(\sum _{i=1}^{\infty }{(-\frac{x}{2})}^i-\sum _{i=1}^{\infty }(-x{)}^i)={\sin }^{-1}(\sum _{i=1}^{\infty }{(-\frac{x}{2})}^i-\sum _{i=1}^{\infty }(-x{)}^i)={\sin }^{-1}(\frac{(-x/2)}{1+\left(x/2\right)}-\frac{(-x)}{1+x})={\sin }^{-1}(\frac{-x}{2+x}+\frac{x}{1+x})$
Equating L.H.S = R.H.S,
$x^2\cdot (\frac{1}{1-x}-\frac{1}{2-x})=x\cdot (\frac{-1}{2+x}+\frac{1}{1+x})\Rightarrow \frac{x^2}{(x^2-3x+2)}=\frac{x}{(x^2+3x+2)}\Rightarrow x^2(x^2+3x+2)=x(x^2-3x+2)\Rightarrow x(x^3+2x^2+5x-2)=0$
So, $x=0$ is one root of the equation.
Now,
$x^3+2x^2+5x-2=0$
$\text{Let}f\left(x\right)=x^3+2x^2+5x-2\therefore f^{\prime }\left(x\right)=3x^2+4x+5\ne 0$
So $f\left(x\right)$ is monotonic in nature.
$f\left(0\right)=-2$ and $f\left(1/2\right)=\frac{9}{8}$
As $f\left(0\right)\cdot f\left(1/2\right)<0$ so one root of $f\left(x\right)$ lies between $0$ and $\frac{1}{2}$.
Hence there are two solutions of the given equation in given interval.