Let's say that there is no limit to $x$, then I believe $y = \arccos(x) + \pi n $ because $\cos(y) = x$ for $x + \pi n , n\in \mathbb{Z}$. However, in the last step of the following calculation, it shows that the solution is $y = \arccos(x) + 2\pi n , y = -\arccos(x) + 2\pi n $:
Is this correct?
On
I'm not sure what your question is. $\cos$ is not injective (it's possible for two different values to have the same output) so there in no inverse function, $f$ where $f(x)= y$ whenever $\cos y = x$.
If your question is to solve all possible $y$ where $\cos y = x$.
We have if $0\le \arccos y \le \pi$ is the only solution between $0$ and $\pi$.
Likewise for any $k\in \mathbb Z$ we have $\arccos y + 2k\pi$ is the only solution between $2k\pi$ and $(2k+1)\pi$.
But what about the solutions between $(2k+1)\pi$ and $(2k+2)\pi$? Or equivalently between $(2k-1)\pi$ and $2k\pi$?
bearing in mind that $\cos(x) = \cos(-x)$ we have $-\pi \le -\arccos y \le 0$ is the only solution between $-\pi$ and $0$ and so $-\arccos y + (2k-1)\pi$ is the only solution between $(2k-1)\pi$ and $2k\pi$.
And thus the solutions will by all the $y$ where $y = \arccos x + 2k\pi$ or $y = -\arccos x + (2k-1)\pi$ for some $k$.
because cos(y)=x for x+πn,n∈Z
That isn't true. if $n$ is an odd number then $\cos(x + \pi n) = -\cos x \ne \cos x$
Take as example $x=1/2$. Then your formula would give the solutions of $\cos y=1/2$ as $y=\text{arccos}(1/2)+\pi n=\pi/3+\pi n$, which in particular would include $y=4\pi/3$ for $n=1$. But $\cos(4\pi/3)=-\cos(\pi/3)=-1/2$, so this is not a solution and your proposed solution set cannot be correct.