<s>[INST] 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?
 [/INST] I use Nghi H Nguyen's method to visually solve trig inequalities by the number unit circle.
Solve: $ F(x) = \cos (3x) < 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) < 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 < 0 $ (blue) that are the 3 open intervals:
$(\frac{\pi}{6}, \frac{\pi}{2})$ and $(5\frac{\pi}{6}, 7\frac{\pi}{6})$ and $(3\frac{\pi}{2}, 11\frac{\pi}{2})$
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