Question

Solve for $x$ and $y$:
$x^2+2x\sin xy+1=0$

Answer 1

$x^2+2xsin\left(xy\right)+1=0⟶\left(1\right)$

We know that, ${\sin }^2\theta +{\cos }^2\theta =1⟶\left(2\right)$

Substituting $\left(2\right)$ in $\left(1\right)$ we get

$\Rightarrow x^2+2xsinxy+{\sin }^{2}xy+{\cos }^{2}xy=0$

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

$\Rightarrow (x^2+2x\sin \left(xy\right)+{\sin }^2\left(xy\right))=({\sin }^2\left(xy\right)-1)$

$\Rightarrow {(x+sin\left(xy\right))}^2=({sin}^2\left(xy\right)-1)$

We see that the left hand side of the above equality is always greater than zero since the square of a quantity is always greater than equal to zero. Whereas the right hand side of the quantity is always than equal to zero.

$\because$ $0\le {sin}^2\left(xy\right)\le 1$

$\Rightarrow -1\le {\sin }^2\left(xy\right)-1\le 0$

For the quantity to hold true we need to have

$x+sin\left(xy\right)=0$ $⟶\left(3\right)$ and ${sin}^2\left(xy\right)=1$

$\Rightarrow$ $sin\left(xy\right)=\pm 1⟶\left(4\right)$

From $\left(4\right)$ we have

$xy=\pi /2$ or $-\pi /2$

For, $xy=\pi /2$

Substituting in $\left(3\right)$ we get

$x+1=0$

$\Rightarrow x=-1$

$\therefore$ $y=-\pi /2$

Similarly for $xy=-\pi /2$

Substituting in $\left(3\right)$ we get, $x-1=0$ $\Rightarrow x=1$

and $y=\pi /2$

$\therefore$ the possible values are $x=-1$ and $y=-\pi /2$ and $x=1$ and $y=\pi /2$