Does the arccos itself contain all solutions or just one solution?

Question

For the equation $$\cos(x)=\frac{1}{2}$$ All solutions are: $$x=\pm\frac{\pi}{3}+p2\pi,\quad p\in\mathbb Z\:.$$

To find these solutions, I use the inverse cosine ($\arccos$ or $\cos^{-1}$). Is the term $p2\pi$ included in the $\cos^{-1}$ itself or is it added to it? More precisely I am asking: Is the solution $\cos^{-1}$ only the $x \in [0,2\pi]$ result or similar (the angle within one cycle) or is it all results?

In other words, which of the following notations are correct:

1. $$\cos(x)=\frac{1}{2}\quad\Leftrightarrow\quad x=\cos^{-1}\left(\frac{1}{2}\right)=\pm\frac{\pi}{3}+p2\pi$$
2. $$\cos(x)=\frac{1}{2}\quad\Leftrightarrow\quad x=\cos^{-1}\left(\frac{1}{2}\right)+p2\pi=\pm\frac{\pi}{3}+p2\pi$$

Answer 1

Generally, the second case is accepted as correct. The reason is that in order for $\cos^{-1}$ to be a function we need there to be a unique output for every input. Therefore we carefully select the domain of definition of $\cos^{-1}$, so that $$\cos^{-1}:[-1,1]\rightarrow[0,\pi].$$ This allows us to treat $\cos^{-1}$ as a well-defined function (in the proper sense of the term), so that, say $\cos^{-1}{(0)}$ has a definite, specified value.

Answer 2

It is not included. The function $\arccos$ is a function and hence one-valued. For $x\in[-1,1]$ it is usually defined as the unique number $t\in[0,\pi]$ with $\cos t=x$. Hence it is not really a (two-sided) inverse function of the cosine: We have $\cos(\arccos(x))=x$ for all $x\in[-1,1]$, and $\arccos(\cos(x))=x$ for all $x\in[0,\pi]$, but certainly not $\arccos(\cos(x))=x$ for all $x\in\mathbb R$.

Also, you are dropping $-\frac\pi 3+p2\pi$ from the solution set.