The problem is finding the real-number $ (x, y) $ of this simultaneous equation $$\begin{cases} (\cos{x})^2+(\sin{y})^2=(x-\dfrac{\pi}{2})^2 \\ (\cos{y})^2+(\sin{x})^2=(y-\dfrac{\pi}{2})^2\end{cases}$$ By adding two equations, I can assume $ (x, y) $is on the circle which is centered at $(\dfrac{\pi}{2},\dfrac{\pi}{2})$ with a radius $\sqrt{2}$, and obviously have the solution when $x=y$. But I can't show the uniqueness of the solution, or find another one. Can anybody help me to solve this problem?


You can solve the circle equation for $y$ and plug it into the first equation, which then gives you

$$ f(x) = (\cos(x))^2 + (\sin(\pi/2 \pm \sqrt{2 - (x-\pi/2)^2 )})^2 - (x-\pi/2)^2 $$ and you are looking for $f(x) = 0$. Here is a plot.

enter image description here

Indeed there are the only solutions $x = y = \pi/2 \pm 1 \simeq \{0.57 ; 2.57\}$. I wouldn't know an easy way of how to prove that formally though. You would need a curve discussion on $f(x)$.


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