how do you find the intersection of two cosine functions

Question

e.mccormick · Answer

Look for when they would be equal.

anonymous · Answer

i know, but how? my trig is rusty. i have 5cosx=5cos2x

e.mccormick · Answer

Well, you can ignore the 5 in front because it is on both.  If you set it to zero and solve, you should find values of x that satify both.  You can then test some to be sure.

anonymous · Answer

how do i factor them though? i have cos2x-cosx=0

e.mccormick · Answer

Know your double angle identity?

anonymous · Answer

yeah but how do i apply that here?

e.mccormick · Answer

They let you get the angle just to x rather than x and 2x.

$\cos(2\theta ) = \cos^2\theta - \sin^2\theta$
$ \text{and }\quad \; = 2cos^2\theta - 1$
$ \text{and }\quad \; = 1 - 2sin^2\theta$

e.mccormick · Answer

Do you see how to factor it if you use the first one?

e.mccormick · Answer

$\cos(2x)-\cos x=0 $

Apply $\cos(2\theta ) = 2cos^2\theta - 1$ to get:

$2\cos^2 x   - 1 - \cos x$

Rearrange it a little:
$2\cos^2 x  - \cos x - 1$
and now it looks like a pretty common quadratic.

The 2 in front is prime, so it will be with only one factor.  The middle is negative, so the number of - cos x is larger than the positive.  The end is -1, so it can only come from a -1 and +1.

(? + 1)(? - 1)  So only thing to figure out is where $\cos x$ and $2\cos x$ go.