How to solve trigonometric equations in 2 variables?

Question

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.

Answer 1

To find two variables you need two equations.

If you had tried $ y = 0 $ you would have get $ x = \pi/2, 3\pi/2... $

Answer 2

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

Answer 3

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 $y=(2n+1)\pi-2x$.