How do we find solutions of cosine function (or sine, etc) for domains outside of arccos? For example:
$$\cos x=1/5$$
$$x \approx1.369, ?, 2\pi(1.369), ?, 4\pi(1.369), ?,...$$
I was taught by several different sources that this is tantamount to asking the question: cosine of what angle(s) is 1/5? Clearly there is an infinite number of solutions. But when using the arccos function we're limited to that function's domain and range. So how do we find the solutions for values of x outside of arccos's domain (the question marks)?
 A: Reasoning geometrically, we get two fundamentally different kinds of solutions for $0 \le x < 2\pi$. 

The first, in the first quadrant, is the $\arccos(1/5) \approx 1.369$ that the arccosine function will give us. But since arccosine is lazy, and refuses to leave its comfort zone in quadrants I and II, we have to supply the value in quadrant IV ourselves. Symmetry considerations tell us this must be $2\pi - \arccos(1/5) \approx 4.913$, since the reference angles above and below the $x$-axis are the same.
But that's two, still not quite the infinitely many you're expecting. The good news is that cosine is as lazy as arccosine: it repeats itself after every $2\pi$ units of input! So if $\arccos(1/5)$ has a cosine of $1/5$, then $2\pi + \arccos(1/5)$ must have the exact same cosine! So to either of these "fundamental" solutions, we can add any integer multiple of $2\pi$ (say, $2\pi k$, where $k$ is any integer) and get yet another angle whose cosine is $1/5$. So, our solution set looks like
$$x = \begin{cases}\arccos(1/5) + 2\pi k,&\text{in quadrant I} \\ 2\pi k - \arccos(1/5), &\text{in quadrant IV}\end{cases},$$
where $k$ can be any integer whatsoever.
A: Draw a cosine curve and use a horizonal line ($y=\frac{1}{5}$ in the example above) on the graph, and you can find all the possible values. The arccos map has the range $[0,\pi]$. To find out all other solutions, just check on your graph how other intersections are related to the one in $[0,\pi]$. 
Specifically, note that $\cos{x}=c$ implies $\cos{(-x)}=c$ and also $\cos{(\pm x+2n\pi)}=c \ (n\in\mathbb{Z})$. Your graph may also show you that those are all the points. 
Therefore, $x$ satisfies $\cos{x}=c$ if and only if
$$x=\pm \arccos{c}+2n\pi,\  n\in\mathbb{Z}.$$
