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 ax^{2}+bx+c)

Though, my teacher helped me get started, and told me to do this:

2cos^{2}(x)-cos(x)-1

**I was lost at how the cos 2(x) became 2cos**^{2}x

Can anyone explain to me this?

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Though, after that I would factor it then find the x's from there.

Re: Find solutions equation in internal [0, 2pi]

Quote:

Originally Posted by

**Chaim** **I was lost at how the cos 2(x) became 2cos**^{2}x

Not just 2cos^{2}x, but 2cos^{2}x - 1. We have 2cos^{2}x - 1 = -(1 - cos^{2}x) + cos^{2}x = cos^{2}x - sin^{2}x = cos 2x.

Re: Find solutions equation in internal [0, 2pi]

Quote:

Originally Posted by

**emakarov** Not just 2cos^{2}x, but 2cos^{2}x - 1. We have 2cos^{2}x - 1 = -(1 - cos^{2}x) + cos^{2}x = cos^{2}x - sin^{2}x = cos 2x.

Ah!

This is the Pythagorean Identities right?

sin^{2}θ+cos^{2}θ=1

Ok thanks!

Re: Find solutions equation in internal [0, 2pi]

The Pythagorean identity **and** the fact that $\displaystyle cos(2x)= cos^2(x)- sin^2(x)$. Do you know that identity?