Question

Number of all pairs $(x,y)$ which satisfy $x^2+2x\sin \left(xy\right)+1=0,\text{if}yϵ(–4\pi ,4\pi )$ is

• A. 4
• B. 8
• C. 10
• D. 6

Answer 1

The correct option is B 8

The equation $x^2+2x\sin \left(xy\right)+1=0$ can be re-written as
$[x+\sin (xy){]}^2+(1–{\sin }^{2}xy)=0$
$\therefore x+\sin \left(xy\right)=0,{\sin }^2\left(xy\right)=1$
${\sin }^2\left(xy\right)=1$ gives $\sin \left(xy\right)=1\text{or}–1$
(i) If $\sin \left(xy\right)=1,\text{then}x+1=0$
$\Rightarrow x=–1$
then $\sin (–y)=1$
or $\sin y=–1$
(ii) If $\sin \left(xy\right)=–1,\text{then}x–1=0$
$\Rightarrow x=1$
then $\sin \left(y\right)=–1$
$\therefore x=\pm 1,\sin y=–1i.e.,y=2n\pi –\frac{\pi }{2},nϵI$
Hence pairs (x, y) are given by $(\pm 1,2n\pi –\frac{\pi }{2}),nϵI$