Finding Intersection of two trig functions

[power09]

The functions are Cos(x) & Sin(2x)
I'm trying to find the area between these two functions at [7pi/2,9pi/2]

All the integration I get but since these lines crossover in this interval I need to find that exact point so I can split up the integrals

I can easily work out that it is between 4pi & 17pi/4 but dont know where to go from there

(the answer is 25pi/6 but I dont know how to find that)

 

[Howard Littlejohn]

So glad I haven't touched a trig function in longer than you've been alive.

If I'm understanding your question, the lines will cross where

cos(x) - sin(2x) = 0

cos(x) - 2 sin(x) cos(x) = 0 [had to look that bit up, aparently sin(2x) = 2sin(x)cos(x)]

cos(x) = 2 sin(x) cos(x)

1 = 2 sin(x)

1/2 = sin(x)

x = what is 30 degrees in radians? pi/6 if 2pi = 360 degrees [vague recollection]

Every 2pi radians is 360 deggrees. So 4pi is the same as zero for trig functions, making your 4pi and 17pi/4 the same as 0 and pi/4. pi/6 sits between those.
Am I even close?

 

[power09]

thanks for replying and yes you are close

I since realized that a equivalent intersection I am looking for obviously occurs near the origin (duh)
then used cos(x)=sin(2x).....sin(x) = 1/2
this occurs at positive pi/6
applied that logic further along the x axis
(4pi + pi/6) = 25pi/6