
Trig Intervals
I'm trying to figure out this problem. I think I have the answer but I'm not sure.
Find all solutions on the interval [0,2$\displaystyle {\pi} $]. When appropriate put you answer in interval notation.
$\displaystyle cos(x)\frac{\sqrt{3}}{2}>0 $
1) this is the same as $\displaystyle cos(x)+0>\frac{\sqrt{3}}{2} $ correct?
2) I know that cosx is equal to $\displaystyle \frac{\sqrt{3}}{2} $ at $\displaystyle \frac{\pi}{6},\frac{11\pi}{6} $
but, since it is asking for greater than, I graphed out cosx and thought that the answer is 0, $\displaystyle {2\pi} $. am I close?

Hi discobob,
What exactly does the graph of cos(x) look like from 0 to 360 degrees?
It starts at cos(0)=1,
drops to zero at 90 degrees,
goes down to 1 at 180 degrees,
comes back up to zero at 270 degrees,
reaches 1 again at 360 degrees.
Therefore there is an interval from zero to where cos(x) is $\displaystyle \frac{\sqrt{3}}{2}$
where $\displaystyle cos(x)>\frac{\sqrt{3}}{2}$
and similar at the other end of the period.

yes, that is the graph i plotted. I think I dont understand how to figure out the interval where $\displaystyle cos(x)>\frac{\sqrt{3}}{2} $. I can see the area where it is greater than $\displaystyle \frac{\sqrt{3}}{2} $, but i dont know how to define it.

Oh! I think I got it.
Is the answer $\displaystyle [0,\frac{\pi}{6}) U (\frac{11\pi}{6},2\pi] $

Note that you need to solve $\displaystyle \cos x\cos \frac{\pi }{6}=2\sin \left( \frac{6x+\pi }{12} \right)\sin \left( \frac{\pi 6x}{12} \right)>0.$
