Solve for x in $ x^2 + y^2 = 1 $ and $ x \pm y = \frac \pi4 $

Question

Solve for x in $ x^2+ y^2 = 1 $ and $ x \pm y = \frac \pi4 $

I tried solving this by substitute method. And using the quadratic formula, but that create lots of cases. The original problem was to solve $ \arcsin x + \arcsin y = \frac\pi2 $ and $ \sin 2x = \cos2y $.

Answer 1

$$\dfrac\pi2\ge\arcsin x=\arccos y\ge0$$

$\implies\dfrac\pi2>1\ge x, y\ge0$ and $x=\sqrt{1-y^2}$

$0\le2x,2y\le2<\pi$

Again, $\cos2y=\sin2x=\cos\left(\dfrac\pi2-2x\right)$

$$2y=2m\pi\pm\left(\dfrac\pi2-2x\right)$$

$+\implies2(x+y)=2m\pi+\dfrac\pi2$

As $0\le2(x+y)<\pi,\implies m=0\implies \dfrac\pi4= x+y=\sqrt{1-y^2}+y$

$\left(\dfrac\pi4-y\right)^2=1-y^2$

Solve for $y\ge0$

Similarly, $-\implies2(y-x)=2m\pi-\dfrac\pi2$

But $2\ge2(y-x)\ge-2,m=?$

Answer 2

Note that $$ (x+y)^2+(x-y)^2=2(x^2+y^2)=2 $$ so $(x\mp y)^2=2-(\pi/4)^2$. This reduces to \begin{align*} x\pm y&=\frac\pi4\\ x\mp y&=\varepsilon\sqrt{2-\frac{\pi^2}{16}} \end{align*} for $\varepsilon=\pm 1$, independent of the $\pm$ in $x\pm y$. Can you solve this and get 4 intersection points?