I need to understand $t=\cos(x) \implies x=\arccos(t)$ I need to understand this because I think that I don't know the meaning of arc...
$t=\cos(x) \Rightarrow x=\arccos(t)$ ??
Thanks
 A: Actually, the truth is slightly more complicated than that. The problem is that if I tell you that $\cos x = t$, I have not yet given you enough informations for you to reconstruct what $x$ is. For example, $x=\frac{\pi}{2}$ and $x=-\frac{\pi}{2}$ both satisfy the equation $\cos x = 0$. In general, therefore, it is not possible to define an inverse of the cosine function.
All is not lost, however, as the cosine function is bijective on the domain $[0,\pi]$. On this interval, it has an inverse, and it is this inverse that we call the $\arccos$ function. This means that, for $x\in [0,\pi]$, you define $\arccos t$ as "the value $x$ for which $\cos x = t$".
This allows you to solve any general equation $\cos x = t$. It immediatelly gives you one solution, $x=\arccos t$. Because $\cos$ is an even function, you immediatelly know that $x=-\arccos t$ is also a solution, as $\cos(-\arccos t)=\cos(\arccos t)=t$. Now, knowing that $\cos$ is a periodic function with period $2\pi$, you also know that if $x$ is a solution, then the value $x+2k\pi$ is a solution for any integer $k$. This means that all solutions are the following:
$$x=\arccos t + 2k\pi, k\in\mathbb Z\\
x=-\arccos t + 2k\pi, k\in\mathbb Z.$$
