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Find all pairs $(x, y)$ with $x, y$ real, satisfying the equations: $\sin\frac{(x+y)}2=0$ & $|x| + |y| = 1$

Working:$\frac{x+y}2=0$ or, $x=-y$

I plotted this.

Plotting $|x| + |y| = 1$, I got a square of side root 2 with x and y-intercept of 1 and -1(different in different quadrants.)

I get $\left(\frac{1}2, \frac{-1}2\right)$ and $\left(\frac{-1}2, \frac{1}2\right)$ I don't know if the points of intersection that I'm getting are correct or not.

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Since you have $x=-y$, it means that $|x|=|y|$. Then you can write $2|x|=1$ or $|x|=\frac 12$. That means $x=\pm\frac12$ and $y=-x$, so your solutions are right. Note that you also need to check if any other solutions exists. $x+y=2k\pi$ with $k\in\mathbb Z$satisfy the first equation. It is easy to see that the only thing that satisfy the second one is $k=0$.

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It is $$\sin(\frac{x+y}{2})=0$$ if $$\frac{x+y}{2}=2k\pi$$ and $$k\in\mathbb Z$$

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