Question

A solution (x, y) of $x^2+2xsinxy+1=0$ is

• A. $(1,0)$
• B. $(1,7\pi /2)$
• C. $(-1,7\pi /2)$
• D. $(-1,0)$

Answer 1

Answer: (B), (C)

$(1,7\pi /2)$
$(-1,7\pi /2)$

$x^2+2xsinxy+1=0\Rightarrow sinxy=\frac{-1-x^2}{2x}$

As $-1\le sin\theta \le 1$

$-1\le \frac{-1-x^2}{2x}\le 1$

This gives $x=-1$ and $x=1$

For $x=-1$

$sin-y=\frac{-2}{-2}\Rightarrow -siny=1\Rightarrow siny=-1\Rightarrow y=(2n+1)\pi +\frac{\pi }{2}$

For $x=1$

$siny=\frac{-2}{2}\Rightarrow y=(2n+1)\pi +\frac{\pi }{2}$

Therefore, for $n=1$

$(x,y)=(1,\frac{7\pi }{2}),(-1,\frac{7\pi }{2})$