I was trying to solve the following pair of equations :
$$ x(y-\sin{y}) = -\pi$$ $$ x(1-\cos{y}) = 2$$
My attempt :
We get, $x \sin{y} = xy + \pi$ and $x \cos y = x - 2$ , then by squaring and eliminating $x^2$ from both sides we get, $x^2 y^2 + 2 xy \pi + {\pi}^2 - 4x + 4 = 0$ and so on. But couldn't solve the equation.
It would also be enough to solve $\sqrt{x} y$ for the problem I am trying . Thanks in advance for help.
I find it easier to use $\sin(x)$ instead of $\sin(y)$ so lets change $y\leftrightarrow x$ to get the system:
$$y=\frac{-\pi}{x-\sin(x)} \tag{1}$$ $$y = \frac{2}{1-\cos(x)} \tag{2}$$
cross multiply noting that $x\neq 2n\pi$, let
$$f(x)=2(x-\sin x)+\pi(1-\cos x ) $$
Where we need to solve $f(x) = 0$. We know for $x>0$, $x \ge \sin(x)$ and $1 \ge \cos(x)$. So for $x > 0$ there are no solutions.
For $x< 0$ analyse the derivative $$\begin{align}f'(x) &= 2-2\cos x +\pi \sin x \\ &= 4 \sin^2(\tfrac{x}{2})+2\pi \sin(\tfrac{x}{2})\cos(\tfrac{x}{2})\\ &= 2\sin(\tfrac{x}{2})(2\sin(\tfrac{x}{2})+\pi \cos(\tfrac{x}{2})) \\ & = 2\sqrt{\pi^2+4}\sin(\tfrac{x}{2})\sin(\tfrac{x}{2}+\arcsin(\tfrac{\pi}{\sqrt{\pi^2+4}})) \end{align}$$
From here for $f'(x) = 0$ we have $x=2n\pi$ or $2n\pi -\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}})$. This provides us with some intervals of monotonicity.
Now for $x \in (-2\pi, -\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}}))$, function is monotonous, and $f(-2\pi) < 0$ and $f(\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}})) > 0$ so we will give only one root, the trivially found $x=-\pi$.
Rewrite $f(x) = 2(x-\sin(x))+2\pi \sin^2(\tfrac{x}{2})$. Here first term is negative for $x < 0$. For $x \in (-\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}}),0)$, $f(x)$ is monotone, but $f(0) = 0$ and $\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}}) > 0$ so no roots.
For $x < -2\pi$, $2x < -4\pi$ but $2\sin(x)-2\pi \sin^2(x)$ can be said to have a lower bound of $-2-2\pi$, that too not achievable. Thus here $2x < 2\sin x -2\pi\sin^2(\tfrac{x}{2})$. So there are no roots here also.
Thus the only root of our system is $(x,y) = (-\pi,1)$ or original equations solutions are $(x,y) = \color{red}{(1,-\pi)}$