<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] Hint:
$$1=(1-\cos y)^2+(1-\sin y)^2=1-2\cos y-2\sin y+1$$ gives you $y$. And at the same time, $2x$.

Alternatively:
$$\frac{\sin 2x+\sin x}{\cos 2x+\cos x}=\frac{2\sin\dfrac{2x+y}2\cos\dfrac{2x-y}2}{2\cos\dfrac{2x+y}2\cos\dfrac{2x-y}2}=\tan\frac{2x+y}2=1$$ so that
$$2x+y=\frac\pi4+k\pi.$$
Then
$$\cos\left(\frac\pi2+2k\pi-y\right)+\cos y=1$$
or 
$$2\cos\left(\frac\pi4+k\pi\right)\cos\left(\frac\pi4+k\pi-y\right)=1.$$
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