<s>[INST] 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. 
 [/INST] Let $z=2x$. Then
$$1=\cos z + \cos y = 2\cos\frac{z-y}2\cos\frac{x+y}2\tag 1$$
$$1=\sin z + \sin y = 2\cos\frac{z-y}2\sin\frac{x+y}2\tag 2$$
Take (1) - (2) to have a factorized equation,
$$\cos\frac{z-y}2 \sin\left( \frac{z+y}2 - \frac\pi4 \right) =0$$
So, two cases to consider:
Case 1) $\cos\frac{z-y}2 = 0$ leads to $z=y + (1+2n)\pi$. Plug it into (1) to see that (1) does not hold. Hence, no solutions.
Case 2) $\sin\left( \frac{z+y}2 - \frac\pi4 \right)=0$ leads to 
$z= -y+\frac\pi2+2n\pi$. Plug it into (1) to get $\cos (y -\frac\pi4) =\frac1{\sqrt2}$ and the solutions for $y$,
$$y=2n\pi, \>\>\>\>\>y =  \frac\pi2 + 2n\pi $$
and the respective $z's$ are $z= \frac\pi2+2k\pi$ and $z= 2k\pi$. 
Thus, there are two sets of the solutions,
$$(x,y) = (\frac\pi4+k\pi, 2n\pi),\>(k\pi, \frac\pi2+2n\pi)$$
</s>