Question

The number of ordered pairs $(x,y)$ satisfying the equation $x^2+2x\sin \left(xy\right)+1=0$ is
$(\text{where}y\in [0,2\pi \left]\right)$

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is B $2$

Given:
$x^2+2x\sin \left(xy\right)+1=0$
$\Rightarrow (x+\sin (xy){)}^2+(1-{\sin }^2\left(xy\right))=0\Rightarrow (x+\sin \left(xy\right){)}^2+{\cos }^2\left(xy\right)=0$
Which is possible only when both terms are individually equal to zero, so

$\Rightarrow x+\sin \left(xy\right)=0\text{and}{\cos }^2\left(xy\right)=0\Rightarrow \sin \left(xy\right)=-x\text{and}\sin \left(xy\right)=\pm 1$

Case $1:\sin \left(xy\right)=1\Rightarrow x=-1$,
$\Rightarrow -\sin \left(y\right)=1\Rightarrow \sin y=-1\Rightarrow y=\frac{3\pi }{2}$

Case $2:\sin \left(xy\right)=-1\Rightarrow x=1$,
$\Rightarrow \sin \left(y\right)=-1\Rightarrow \sin y=-1\Rightarrow y=\frac{3\pi }{2}$

Hence, there are $2$ ordered pairs $(-1,\frac{3\pi }{2})$ and $(1,\frac{3\pi }{2})$.