At some point when dealing with a problem I had to solve the following system $$\begin{cases} \cos(x)-\cos(x+y)=0\\ \cos(y)-\cos(x+y)=0 \end{cases}$$ I seem to have problems with doing so, or maybe just with writing the answer in the simplest possible form.

The system is equivalent to $$\begin{cases}-2\sin\frac{x+x+y}{2}\sin\frac{x-x-y}{2}=0\\ -2\sin\frac{y+x+y}{2}\sin\frac{y-x-y}{2}=0 \end{cases} \iff \begin{cases}\sin\frac{2x+y}{2}\sin\frac{y}{2}=0\\ \sin\frac{x+2y}{2}\sin\frac{x}{2}=0\end{cases}$$ The first equation gives $\dfrac{2x+y}{2}=k\pi$ or $\dfrac{y}{2}=m\pi$, and the second one gives $\dfrac{x+2y}{2}=n\pi$ or $\dfrac{x}{2}=l\pi,$ where $k, m, n, l\in\mathbb{Z}$. So a set of solutions would come from $$\begin{cases}\dfrac{x+2y}{2}=n\pi \\ \dfrac{2x+y}{2}=k\pi \end{cases}\iff \begin{cases}x+2y=2n\pi \\ 2x+y=2k\pi \end{cases},$$ so $-2x-4y=-4n\pi$ (multiplying the first equation by $(-2)$ and adding to the second). This gives $$-3y=-2k\pi-4n\pi=\pi(2k-4n)=\pi.p,$$ where $p$ is even.

So we have $x=2n\pi+\dfrac{2}{3}p\pi=2n\pi+\dfrac{1}{3}p\pi$ (remember $p$ was even) and $y=-\dfrac{1}{3}p\pi$. I am pretty sure this isn't the simplest form giving this set of solutions of the system. Any hints on how I could write it (solve it) better would be appreciated! Thanks!

## Best Answer

The equations can be solved more easily than the OP proposes. From the properties of the cosine function:

$$\pm x = x+y - 2k\pi$$

$$\pm y = x + y - 2m\pi$$

It turns out that only the cases with a minus sign lead to something useful:

$$2x + y = 2k\pi$$

$$2y + x = 2m\pi$$

Eliminating as usual $x$ or $y$, we arrive at $x = \frac {2}{3}\pi (2k-m)$ and $y = \frac {2}{3}\pi (2m-k)$. Where $k$ and $m$ are natural numbers. If we limit the solutions to the interval $(0,2\pi)$ we obtain these three: $x = y = 0$, $x = y = \frac{2}{3}\pi$ and $x=y = \frac{4}{3}\pi$.