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

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.

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