The number of real solution of
$\displaystyle \sin^{-1}\bigg(\sum^\infty_{i=1}x^{i+1}-x\sum^\infty_{i=1}\bigg(\frac{x}{2}\bigg)^i\bigg)=\frac{\pi}{2}-\cos^{-1}\bigg(\sum^\infty_{i=1}\bigg(-\frac{x}{2}\bigg)^i-\sum^\infty_{i=1}(-x)^i\bigg)$ lying in $(-1,1)$
(Here $\displaystyle \sin^{-1}\in\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]$ and $\displaystyle \cos^{-1}\in[0,\pi])$
What I try : $\displaystyle \sin^{-1}\bigg(\frac{x^2}{1-x}-x\frac{\frac{x}{2}}{1-\frac{x}{2}}\bigg)+\cos^{-1}\bigg(\frac{-\frac{x}{2}}{1+\frac{x}{2}}-\frac{(-x)}{1+x}\bigg)=\frac{\pi}{2}$
(Above i have used sum of infinite geometric progression )
So we have $\displaystyle \frac{x^2}{1-x}-\frac{x^2}{2-x}=\frac{x}{1+x}-\frac{x}{2+x}$
$\displaystyle x=0$ or $\displaystyle \frac{x}{(1-x)(2-x)}=\frac{1}{(1+x)(2+x)}$
So one solution is $x=0$ but I am struck in other real solution.
Now how do I solve it, help me please
Best Answer
I think you have correctly got
Now, $$\frac{x}{(1-x)(2-x)}=\frac{1}{(1+x)(2+x)}$$ is equivalent to $$x(1+x)(2+x)=(1-x)(2-x)$$ i.e. $$x^3+2x^2+5x-2=0$$ Let $f(x)$ be the LHS.
Then, since $f'(x)=3x^2+4x+5\gt 0$, we see that $f(x)$ is strictly increasing with $$f(-1)=-6\lt 0\qquad\text{and}\qquad f(1)=6\gt 0$$ So, we see, by the intermediate value theorem, that $f(x)=0$ has exactly one real solution $x=\alpha$ in $(-1,1)$.
Here, note that $\alpha\not=0$ since $f(0)=-2\not=0$.
Therefore, the number of the real solutions of the given equation is $\color{red}2$.