If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$ 
*

*$\sin x + \sin y = 1$

*$\cos x + \cos y = 0$


Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.

How can I solve this equation?

$$\sin x + \sin y = 1\Rightarrow 2\sin\frac{x+y}{2}\cos \frac{x-y}{2}=1\\\cos x+\cos y=0\Rightarrow 2\cos\frac{x+y}{2}\cos \frac{x-y}{2}=0$$
It follows $$ \cos \frac{x+y}{2}=0\text{ and } \cos \frac{x-y}{2}=0\iff  $$ The first ($\cos \frac{x+y}{2}=0$) gives, from the first given equation,
$$\begin{cases}x+y=\pi\\x-y=\frac{2\pi}{3}\end{cases}\iff(x,y)=(\frac{\pi}{6},\frac{5\pi}{6})$$
The second  ($\cos \frac{x-y}{2}=0$) is discarded by incompatibility.
Thus the solutions are given by $$(x,y)=( \frac{\pi}{6}+2m\pi,\space \frac{5\pi}{6}+2n\pi)$$
It is not worthless to see the graphic solutions: all intersections of the red lattice, with blue closed curves.

