If I have a relation: $$R=\{(x,y)\in\mathbb R^2 : \cos(x)=\cos(y)\},$$ it is clear to me that $[x]= \{x,-x : x \in \mathbb R\}$
What I'm trying to say is that the equivalence class $x$ is partitioned into the set of all $x$ with $-x$. It is clear because of course $\cos(x)=\cos(-x)$ OR $\cos(x)=\cos(x)$, and of course $\cos(x)$ is defined for all real numbers.
I am just unsure of how this would be written correctly. Thank you for your assistance in advance.
The correct way to write that the equivalence class of $x$ consists of $x$ and $-x$ is to say $[x]=\{x,-x\}$ for each $x\in\mathbb{R}$. However, as noted in thanasisddr's answer, the equivalence class of $x$ actually consists of elements of the form $\pm x+2k\pi$ for $k\in \mathbb{Z}$. You can write this as $$[x]=\{x+2k\pi:k\in\mathbb{Z}\}\cup\{-x+2k\pi:k\in\mathbb{Z}\}.$$
If you want to write the entire partition as a set, it would then be $P=\{[x]:x\in\mathbb{R}\}$, where $[x]$ is defined as above, or more directly $$P=\{\{x+2k\pi:k\in\mathbb{Z}\}\cup\{-x+2k\pi:k\in\mathbb{Z}\}:x\in\mathbb{R}\}.$$ That is, $P$ is the set of all expressions of the form $\{x+2k\pi:k\in\mathbb{Z}\}\cup\{-x+2k\pi:k\in\mathbb{Z}\}$, where $x$ can be any element of $\mathbb{R}$.