Given to find $\arccos\left(\cos(\frac{14 \pi}{3})\right)$ Okay so $\arccos\left(\cos\left(\dfrac{14 \pi}{3}\right)\right)$ can be written as 
$\arccos\left(\cos\left(4 \pi+\dfrac{2 \pi}{3}\right)\right)$ yielding answer as $\frac{2\pi}{3}$
But why can't we take it the following way $\arccos\left(\cos\left(5 \pi-\dfrac{\pi}{3}\right)\right)$???
 A: The function $x \mapsto \cos x$ is not one-to-one, for example $\cos(0) = 1$ and $\cos(2\pi) = 1$.
That means that there is no inverse function. What is $\arccos(0)$? Well, $\cos(2\pi k)=1$ for all integers $k$.
We usually restrict the domain of cosine to make it a one-to-one function. Usually, we say that $0 \le x \le \pi$. This makes $x \mapsto \cos x$ a one-to-one function, where $\cos : [0,\pi] \to [-1,1]$.
The inverse function is given by $\arccos : [-1,1] \to [0,\pi]$.
Having said that, there are lots of different domains that make cosine one-to-one, for example $6\pi \le x \le 7\pi$ or $-13\pi \le x \le -12\pi$.
The function cosine makes sense for any input, so $\cos(1234567)$ makes sense, but the output will be in the interval $[-1,1]$. It depends on your choice of domain to make sense of the inverse of a value in the interval $[-1,1]$. As I said:
$$\{ x \in \mathbb R : \cos x = 0\} = \{  2\pi k : k \in \mathbb Z\}$$
A: Remember that $$\arccos(\cos(x)) = x \iff x\in[0,\pi]$$
so
$$\arccos\bigg(\cos(\dfrac{14 \pi}{3})\bigg) =\arccos\bigg(\cos(\dfrac{14 \pi}{3}-4\pi)\bigg)=\arccos\bigg(\cos(\dfrac{2 \pi}{3})\bigg)= \dfrac{2 \pi}{3}$$
