Solve these equations simultaneously (trig) Solve for $ x,y: $ 

\begin{equation}\cos x -\cos(x+y) = 0
\end{equation}

\begin{equation}\cos y -\cos(x+y) = 0
\end{equation}

The answers are $(0, 0), (\frac{2\pi}{3}, \frac{2\pi}{3})$. 

I get $(0, 0)$, but how do you get the latter?

Sorry if this is a bit basic - I don't remember trig well...

tfa
Given

$$\cos x-\cos(x+y)=0\\
\cos y-\cos(x+y)=0$$

Subtracting the second equation from the first yields

$$\cos x=\cos y$$

Recall that $\cos x=\pm\sqrt{1-\sin^2x}$ and $\cos y=\pm\sqrt{1-\sin^2x}$.
This now becomes $\sqrt{1-\sin^2x}=\pm\sqrt{1-\sin^2y}$ yielding

$$\sin x=\pm\sin y$$

Recall that $\cos(x+y)=\cos x\cos y-\sin x\sin y$.
Thus, $\cos x-\cos(x+y)=0$ now becomes $\cos x-\cos^2x\pm\sin^2x=0$.
Solving $\cos x-\cos^2x-\sin^2x=0$ yields $\cos x=1$.
Solving $\cos x-\cos^2x+\sin^2x=0$ yields $\cos x-\cos^2x+1-\cos^2x=0$ which is equivalent to

$$(2\cos x+1)(-\cos x+1)=0$$

This yields two possibilities:  $\cos x=1$ or $\cos x=-\frac12$.  Can you do the rest?
I should begin dividing first equation by cos(x), then second by cos(y), ....