how to solve trigonometric inequalities? how does one solve trigonometric inequalities? Is there a method to this or is every solution done ad hoc? 

simple equations of the type: $cos3x \leq  0$  when: $0\leq x \leq 2π$

The attempt at a solution: equating $cos 3x = 0$  yields $$ π /6  + 2\frac13πk\leq x \leq 2π -π /6 - πk /3 $$ as a general solution...what happens next?
There is a sort of general method. Take your example: $\cos 3x \leq 0$ for $0 \leq x \leq 2 \pi$.

Like you did, first solve the equality $\cos 3x = 0$. You cannot always do this exactly, but here we can. We get 
$$
3x = \frac{1}{2}\pi + 2 k \pi \quad\vee\quad 3x = \frac{3}{2} \pi + 2 k \pi
$$. By drawing a graph (or looking at the derivative of $\cos 3x$ at those points), we know that the inequality holds for $3x$ between them. So $$
\frac{1}{2}\pi + 2 k \pi \leq 3x \leq \frac{3}{2}\pi + 2 k \pi
$$
It is important here that you get a representation for the interval. In your example solution, you gave representations for the endpoints of the interval with different meanings for $k$ on the left-hand side and the right-hand side. Your thing works for $k = 0$, but for $k = 10$ the left endpoint is to the right of the right endpoint.

Dividing by 3 we get
$$
\frac{1}{6}\pi + \frac{2}{3} k \pi \leq x \leq \frac{1}{2}\pi + \frac{2}{3} k \pi
$$

This is a general solution for all of the real line. To confine ourselves to values of $x$ in $[0, 2\pi]$, we try different values of $k$ and see if the resulting interval lies within $[0, 2\pi]$. In this case, that holds for $k = 0, 1, 2$.

We get:
$$
k = 0 \;\rightarrow\; x \in [\frac{1}{6} \pi, \frac{1}{2} \pi] \\
k = 1 \;\rightarrow\; x \in [\frac{5}{6} \pi, \frac{7}{6} \pi] \\
k = 2 \;\rightarrow\; x \in [\frac{3}{2} \pi, \frac{11}{6} \pi]
$$

The answer will be the union of these three intervals.
I use Nghi H Nguyen's method to visually solve trig inequalities by the number unit circle.
Solve: $ F(x) = \cos (3x) &lt; 0$
There are totally 6 end points at: $ \frac{\pi}{6}, \frac{\pi}{2}, 5\frac{\pi}{6}, 7\frac{\pi}{6}, 3\frac{\pi}{2} $ and $11\frac{\pi}{6} $that divide the unit circle into $6$ equal arc lengths.
Use the check point method to find the sign status of arc length $(\frac{\pi}{6}, \frac{\pi}{2})$.
Select $x = \frac{\pi}{3}.$
We get : $F(x) = \cos 3x = \cos \pi = -1. $Then, $f(x) &lt; 0 $ inside this arc length.
Color it blue. Use endpoint's property to easily find other sign status of $F(x)$ for other arc lengths.
We have, starting from end point $\frac{\pi}{6}$ : Blue, red, blue, red, blue, red.
We get the solution set of $F(x) = \cos 3x &lt; 0 $ (blue) that are the 3 open intervals:
$(\frac{\pi}{6}, \frac{\pi}{2})$ and $(5\frac{\pi}{6}, 7\frac{\pi}{6})$ and $(3\frac{\pi}{2}, 11\frac{\pi}{2})$