How can I solve this system of trigonometric equations for $0 \leq a\leq 2\pi$ and $0\leq b\leq2\pi$
$$\cos a + \cos (a+b) = 0$$
$$\cos b + \cos (a+b) = 0$$
the plot shows three points of intersection: http://www.wolframalpha.com/input/?i=ContourPlot%5B%7BCos%5Bx%5D+%2B+Cos%5Bx+%2B+y%5D+%3D%3D+0%2C+Cos%5By%5D+%2B+Cos%5Bx+%2B+y%5D+%3D%3D+0%7D%2C+%7Bx%2C+0%2C2+pi%7D%2C+%7By%2C+0+%2C+2pi%7D%5D
$$\cos(A+B)=-\cos A=\cos(\pi-A)\implies A+B=2m\pi\pm(\pi-A)$$ where $m$ is any integer
Taking the - sign, $B=(2m-1)\pi$
$$0=\cos B+\cos(A+B)=\cos((2m-1)\pi)+\cos(A(2m-1)\pi)$$ $$\iff\cos A=1\implies A=2r\pi$$
Consider the + sign similarly.
$$B=(2m+1)\pi-2A$$
$$0=\cos B+\cos(A+B)=\cos\{(2m+1)\pi-2A\}+\cos\{A+(2m+1)\pi-2A\}$$ $$\implies-\cos2A-\cos A=0$$
$$\implies2\cos^2A+\cos A-1=0$$