Solution for $x$, $\sin^{-1} (x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots)+\cos^{-1} (x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\frac{\pi}{2}$

Question

Solution for $x$, $\sin^{-1} (x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots)+\cos^{-1} (x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\frac{\pi}{2}$

I have tried,

$ 1-\frac{x^2}{2}+\frac{x^3}{4}-\cdots=:y$

Now, rewrite the equation in terms of $y$,

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

$\Rightarrow x-1+y-x^2y=0\Rightarrow(x-1)(1+y+xy)=0$

But, the answer is given $0$,$1$ . How to get $x=0$ solution from my process?(I admits that $0$ is a root of the equation.)

Answer 1

$x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots$ is not equal to $x^2y:=x^2\left(1-\frac{x^2}{2}+\frac{x^3}{4}-\cdots\right).$

Instead, define, for every $x\in D:=(-2,2):$ $$f(x):=x\left(1-\frac x2+\frac{x^2}4-\dots\right)=x\sum_{n=0}^\infty\left(\frac{-x}2\right)^n=\frac x{1+\frac x2},$$ Note that $|f(x)|\le1$ iff moreover, $x\ge-\frac23.$

The equation thus only makes sense for $-\frac23\le x<\sqrt2,$ and rewrites then $$\arcsin f(x)+\arccos f(x^2)=\frac\pi2,$$

which is equivalent to $$f(x)=f(x^2).$$ Since $f$ is one-to-one (on $D$), the solutions are $0$ and $1.$