Difference in solving a trigonometric equation and evaluating trig functions I just come to be curious about answering this question. It may look like basic, but explanation(s) like showing off difference in solving equation and evaluating trig. functions weren't found in any of the text I read.
$$\ x= arctan(\frac{-\sqrt3}{3})$$
where the answer is $\ x = \frac{5\pi}{6}+ \pi n$
However, finding out that if we look for $\ arctan(\frac{-\sqrt3}{3})$ - the answer is $\frac{-\pi}{6}$ because rng: $\frac{-\pi}{2}$< x < $\frac{\pi}{2}$ 
Is it because with equations, we find all sets of possible solutions while evaluation we follow what is the indicative ranges for a specific trig functions?
P.S. Never have been taught in school properly but only have study for myself. Hope you understand my situation.
"edit: I apologize for the typo earlier, it should be $\ x= arctan(\frac{-\sqrt3}{3})$
 A: It depends on how you define the arctangent. The most common definition is

for $y$ a real number, then $\arctan y$ denotes the unique number $x\in(-\pi/2,\pi/2)$ such that $\tan x=y$.

With this definition, $\arctan(-\sqrt{3}/3)$ (I assume you have a typo) is not $5\pi/6$, but $-\pi/6$.
Most likely, you have an elementary equation in the tangent
$$
\tan x=a
$$
Then the solutions of the equation are all numbers of the form
$$
x=\arctan a+\pi n
$$
Of course, writing the solutions in the form
$$
x=\pi+\arctan a+\pi n
$$
is exactly the same thing.

I have seen textbooks that define the arctangent as a multivalued function. In this case, writing $-\pi/6+\pi n$ or $5\pi/6+\pi n$ makes no difference, because
$$
\left\{-\frac{\pi}{6}+\pi n: n\in\mathbb{Z}\right\}
=
\left\{\frac{5\pi}{6}+\pi n: n\in\mathbb{Z}\right\}
$$
I'm inclined to think it's not your case.
A: First, there is a typo, as $\sin\frac\pi6=\frac12\implies\cos\frac\pi6=\frac{\sqrt3}2$ so that
$$
\tan(-\frac{\pi}6)=-\frac1{\sqrt3}=-\frac{\sqrt3}3
$$

Even with correction, the first formulation is wrong. The set $\{\frac{5\pi}6+k\pi:k\in\Bbb Z\}$ is the solution to the equation $$\tan x=-\frac{\sqrt3}3.$$
The assignment
$$
x=\arctan(-\frac{\sqrt3}3)
$$
indeed demands the evaluation of the main branch of the inverse tangent with result $-\frac{\pi}6$.
A: $\tan(x) $ is periodic and so not one-to-one; $\tan\left(\dfrac{5\pi}6+\pi n\right)=\dfrac{-1}{\sqrt3}$ for all integers $n$.  
When only one value is desired for $\arctan\left(\dfrac{-1}{\sqrt3}\right)$, it may be restricted to the principal value $\dfrac{-\pi}6$, 
which is between $\dfrac{-\pi}2$ and $\dfrac{\pi}2$.
