Solve a system of equation: $\cos(2x) + \cos(y) = 1$, $\sin(2x) + \sin(y) = 1$ Solve a system of equation:
$$\cos(2x) + \cos(y) = 1$$
$$\sin(2x) + \sin(y) = 1$$
My idea:
Let's see what is product of this two equations.
$$\cos(2x)\sin(2x) + \cos(2x)\sin(y) + \cos(y)\sin(2x) + \cos(y)\sin(y) = 1$$
$$\cos(2x)\sin(2x) + \sin(2x+y) + \cos(y)\sin(y) = 1$$ But this idea didn't give me anything. Also if I sum I have problem... but this is high school problem so it must have some easy solution. 

HINT.-We have from the two given equations $$2\cos\left(\dfrac{2x+y}{2}\right)\cos\left(\dfrac{2x-y}{2}\right)=1\\2\sin\left(\dfrac{2x+y}{2}\right)\cos\left(\dfrac{2x-y}{2}\right)=1$$ so, by division $$\tan\left(\dfrac{2x+y}{2}\right)=1\Rightarrow\dfrac{2x+y}{2}=\dfrac{(2n+1)\pi}{4}$$ Each of the two given equations is an "oval" closed curve over some bounded region of the plan and this is repeated periodically because of the periodicity of functions considered. 
Now taking $n=0$ the line $y=-2x+\dfrac{\pi}{2}$ cuts the axes in $(0,\dfrac{\pi}{2})$ and
$(\dfrac{\pi}{4},0)$ which give points of intersection satisfying the proposed equations. Thus the general solution, taked from paralelle lines to the first one is $$(x,y)=\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{5\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{9\pi}{4},\dfrac{\pi}{2}\right),\cdots,\left(\dfrac{(4n+1)\pi}{4},\dfrac{\pi}{2}\right),\cdots$$
