Question

Number of value(s) of $x$ for which ${\sin }^{-1}(x^2-\frac{x^4}{3}+\frac{x^6}{9}...)+{\cos }^{-1}(x^4-\frac{x^8}{3}+\frac{x^{12}}{9}...)=\frac{\pi }{2}$, where $0\le \left|x\right|<3^{1/4}$, is

Answer 1

${\sin }^{-1}(x^2-\frac{x^4}{3}+\frac{x^6}{9}-...)+{\cos }^{-1}(x^4-\frac{x^8}{3}+\frac{x^{12}}{9}...)=\frac{\pi }{2}\Rightarrow (x^2-\frac{x^4}{3}+\frac{x^6}{9}.....)=(x^4-\frac{x^8}{3}+\frac{x^{12}}{9}......)\Rightarrow \frac{3x^2}{3+x^2}=\frac{3x^4}{3+x^4}\Rightarrow x^2=x^4\Rightarrow x^2=0,1\therefore x=-1,0,1$
Hence, the number of values is equal to $3.$