Question

The number of pair solution (x, y) of the equation $1+x^2+2xsin\left(co{s}^{-1}y\right)=0$ is ___ .

Answer 1

$\sin \left({\cos }^{-1}y\right)=\frac{-(x+\frac{1}{x})}{2};x\ne 0$

Only possibility is

$\sin {\cos }^{-1}y=\pm 1$ & hence x $=\pm 1$

if $x=1\Rightarrow \sin \left({\cos }^{-1}y\right)=-1$

$\Rightarrow$ Not Possible $[\because 0\le {\cos }^{-1}y\le \pi ]$

if $x=-1\Rightarrow \sin \left({\cos }^{-1}y\right)=1$

$\Rightarrow y=0$

So $x=-1$ & $y=0$ is solution