Question

The number of paired solutions $(x,y)$ of the equation $1+x^2+2x\sin \left({\cos }^{–1}y\right)=0$ is

Answer 1

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

Only possibility is

$\sin \left({\cos }^{-1}y\right)=\pm 1$ & 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 the solution.