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. 
 A: Hint
$$\cos2x-\sin2x=\sin y-\cos y$$
$$\cos(2x+45^\circ)=-\cos(y+45^\circ)=\cos(y-135^\circ)$$ as $\cos(z-180^\circ)=-\cos z$
$$2x+45^\circ=360^\circ n\pm(y-135^\circ)$$ where $n$ is any integer
Considering $+$ sign,
$$2x=360^\circ n+ y-180^\circ$$
For $-$ sign, $$2x=360^\circ n+90^\circ-y$$
Hope you can take it home from here?
A: 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 
$$2\cos\left(\frac\pi4+k\pi\right)\cos\left(\frac\pi4+k\pi-y\right)=1.$$
A: 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,
$$(x,y) = (\frac\pi4+k\pi, 2n\pi),\>(k\pi, \frac\pi2+2n\pi)$$
A: $Cos (2x)+Cos(y)=1$ ⇒$Cos (2x)=1-Cos(y)$⇒ $Cos^2(2x)=1+Cos^2(y)-2Cos(y)$
$Sin(2x)+Sin(y)=1$ ⇒$Sin^2x=1+Sin^2(y)-2Sin(y)$
Summing these relations we get:
$Sin (y)+Cos (y)=1$ ⇒ $y=(2k+1)\frac{\pi}{2}$, or $y=2k\pi$
If $y=(2k+1)\frac{\pi}{2}$, then $Cos(2x)=1$⇒ $2x=2k\pi$⇒$x=k\pi$
If $y=2k\pi$, then $Cos (2x)=0$⇒$2x=(2k+1)\frac{\pi}{2}$ ⇒ $x=(2k+1)\frac{\pi}{4}$ 
A: You could start by computing $\cos^2 2x+\sin^2 2x$ to eliminate the terms in $2x$, for example.
When you start manipulating equations like this you are trying to simplify in some way. So what were you trying to do by multiplying them together? Start thinking about how to spot a potentially good idea, rather than just an idea.
A: 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 $$(x,y)=\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{5\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{9\pi}{4},\dfrac{\pi}{2}\right),\cdots,\left(\dfrac{(4n+1)\pi}{4},\dfrac{\pi}{2}\right),\cdots$$
