# Find solutions equation in internal [0, 2pi]

• May 8th 2012, 03:15 PM
Chaim
Find solutions equation in internal [0, 2pi]
cos 2x = cos x

Well first thing I would have to do is convert this into a quadratic formula (meaning ax2+bx+c)
Though, my teacher helped me get started, and told me to do this:

2cos2(x)-cos(x)-1
I was lost at how the cos 2(x) became 2cos2x

Can anyone explain to me this?

----
Though, after that I would factor it then find the x's from there.
• May 8th 2012, 03:46 PM
emakarov
Re: Find solutions equation in internal [0, 2pi]
Quote:

Originally Posted by Chaim
I was lost at how the cos 2(x) became 2cos2x

Not just 2cos2x, but 2cos2x - 1. We have 2cos2x - 1 = -(1 - cos2x) + cos2x = cos2x - sin2x = cos 2x.
• May 8th 2012, 05:04 PM
Chaim
Re: Find solutions equation in internal [0, 2pi]
Quote:

Originally Posted by emakarov
Not just 2cos2x, but 2cos2x - 1. We have 2cos2x - 1 = -(1 - cos2x) + cos2x = cos2x - sin2x = cos 2x.

Ah!
This is the Pythagorean Identities right?
sin2θ+cos2θ=1

Ok thanks!
• May 9th 2012, 10:42 AM
HallsofIvy
Re: Find solutions equation in internal [0, 2pi]
The Pythagorean identity and the fact that $cos(2x)= cos^2(x)- sin^2(x)$. Do you know that identity?