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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?

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I'm afraid if you don't provide a couple examples, or further narrow your question, this may be closed by some users as "too broad," or as "missing context". –  amWhy Nov 20 '13 at 15:13
    
thanks, Ill provide an example shortly... –  Bak1139 Nov 20 '13 at 15:14
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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.

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Following the way you suggested, my only final solution here was $\frac{1}{6}\pi \leq x \leq \frac{11}{6}\pi $. Is this ok? –  Bak1139 Nov 23 '13 at 19:09
    
No, it's more involved than that. To get a feel for these sort of situations, plots are extremely helpful as well to check your answer. I edited my answer a bit, is this more clear? –  Mark Nov 25 '13 at 18:45
    
Yeah, I think I got it, thanks man. –  Bak1139 Nov 27 '13 at 14:24
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