If I am corrrect, the equation defining the inverse of a function $f$ can be found by switching $x$ and $y$.
Consequently, I expected $x = cos(y)$ to have the same graph as $y=cos^{-1} (x)$.
How to explain it's not actually the case?
Which one qualifies as the inverse of the function $cos(x)$?
The reason is that $\cos^{-1}$, despite the name, is not the inverse of $\cos$. Remember, $\cos$ is not actually invertible, since $\cos$ is not one-to-one. The graph of $x = \cos(y)$ shows the curve failing the vertical line test, rather dramatically.
Instead, $\cos^{-1}$ is a right-inverse of $\cos$, meaning that $\cos(\cos^{-1}(x)) = x$ for all applicable $x$ (i.e. $x \in [-1, 1]$). On the other hand, it is not a left inverse, in the sense that it is not necessarily the case that $\cos^{-1}(\cos(x)) = x$ for all $x \in \Bbb{R}$. As a concrete example, $$\cos^{-1}(\cos(2\pi)) = \cos^{-1}(1) = 0 \neq 2\pi.$$
The reflection rule applies to actual inverses, which need the function to be one-to-one (and onto, although you can always restrict the codomain to the range). One could say that $x = \cos(y)$, while it doesn't define a function of $y$ in terms of $x$, it is the inverse relation to $\cos$, which allows for a one-to-many relationship.