<s>[INST] If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$ 
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*$\sin x + \sin y = 1$

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


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?
 [/INST] I will only look for solutions on $[0, 2\pi)$
We need to make $2$ observations:
$1)$ The maximum value of $\sin \theta$ is $1$. Therefore, if any of $\sin x$, $\sin y$ is negative, $\sin x + \sin y < 1$. It follows that both $x$ and $y$ must be $\in [0, \pi]$. 
$2)$ $\cos x= -\cos y \iff \cos x = \cos (\pi - y)$ . (You can prove this using the cosine subtraction formula). Since we are working only on the interval $[0, \pi]$, we must have $x=\pi-y$. This makes sense, because the cosines of symmetric angles on opposite sides of the $y$ axis will cancel out to $0$.

Substitute $x=\pi-y$ into the first equation:
$$\sin (\pi-y) + \sin y = 1$$
Using $\sin (a-b)= \sin a \cos b - \cos a \sin b$ we can deduce that $\sin (\pi-y)=\sin y$
$$\sin y + \sin y = 1$$
$$\sin y = \dfrac 12$$
$$y = \dfrac {\pi}{6}, y= \dfrac {5\pi}{6}$$
$$x=\pi-y$$
Therefore the solutions are $ \left( \dfrac {\pi}{6}, \dfrac {5\pi}{6}  \right)$ and $ \left(\dfrac {5\pi}{6}, \dfrac {\pi}{6}  \right)$. The symmetry of the solutions is expected, since both original equations and symmetric.
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