How to solve trigonometric equations in 2 variables? Find out both $x$ and $y$, $$\cos(x)=-\cos(x+y)$$ I come up with this equation when I was finding out maxima and minima of a two variable function $$f(x,y)=\sin x+\sin y+\sin (x+y);$$ however I get a solution by hit and trial approach, that is $$x=y={\pi\over 3}$$ will satisfy this equation, but how to solve it, as I have many other problems of the same kind and this hit and trial is time consuming.
 A: I would try $x=a+b$, $x+y=a-b$. Then
$$\begin{align}\cos x&=\cos(a+b)=\cos a\cos b-\sin a\sin b\\
&=-\cos(x+y)=-\cos(a-b)=-\cos a\cos b-\sin a\sin b\end{align}$$
So that brings us to $2\cos a\cos b=0$ so either $\cos a = 0$ in which case $a=\left(n+\frac12\right)\pi$, $y=(a-b)-(a+b)=-2b$, so $x=\left(n+\frac12\right)\pi-\frac12y$ or $\cos b=0$ so $b=\left(n+\frac12\right)\pi$, $y=-(2n+1)\pi$, and $x=a-(2n+1)\pi$.
Summarizing, we could have $y=\text{anything}$ but then $x=\left(n+\frac12\right)\pi-\frac12y$ or $x=\text{anything}$ but $y=(2m+1)\pi$.
EDIT: I guess another way of expressing the solution would be to say that $x=\text{anything}$ and $y=(2n+1)\pi$ or $y=(2n+1)\pi-2x$.
A: To find two variables you need two equations.
If you had tried $ y = 0 $ you would have  get $ x = \pi/2,  3\pi/2... $ 
A: On the top of your solution, finding $x$ and $y$ for: $$Cos(x)=-Cos(x+y)$$
I. If  $y = 0$ then any x$\in$ $\Re$ is a solution.
II. By the trigonometric identity $Cos(x)=-Cos(x+\pi)$ we get that if $y = \pi$,  x can be again any real number.
To solve these type of problems in general, I would look through the identities first and see if any of them are similar to my problem. Identities
