# Trig Intervals

• January 18th 2010, 06:33 AM
discobob
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 ${\pi}$]. When appropriate put you answer in interval notation.

$cos(x)-\frac{\sqrt{3}}{2}>0$

1) this is the same as $cos(x)+0>\frac{\sqrt{3}}{2}$ correct?
2) I know that cosx is equal to $\frac{\sqrt{3}}{2}$ at $\frac{\pi}{6},\frac{11\pi}{6}$

but, since it is asking for greater than, I graphed out cosx and thought that the answer is 0, ${2\pi}$. am I close?
• January 18th 2010, 06:52 AM
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 $\frac{\sqrt{3}}{2}$

where $cos(x)>\frac{\sqrt{3}}{2}$

and similar at the other end of the period.
• January 18th 2010, 09:42 AM
discobob
yes, that is the graph i plotted. I think I dont understand how to figure out the interval where $cos(x)>\frac{\sqrt{3}}{2}$. I can see the area where it is greater than $\frac{\sqrt{3}}{2}$, but i dont know how to define it.

• January 18th 2010, 10:07 AM
discobob
Oh! I think I got it.

Is the answer $[0,\frac{\pi}{6}) U (\frac{11\pi}{6},2\pi]$
• January 18th 2010, 10:21 AM
Krizalid
Note that you need to solve $\cos x-\cos \frac{\pi }{6}=2\sin \left( \frac{6x+\pi }{12} \right)\sin \left( \frac{\pi -6x}{12} \right)>0.$
• January 18th 2010, 10:26 AM