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Math Help - Find solutions equation in internal [0, 2pi]

  1. #1
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    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.
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  2. #2
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    Re: Find solutions equation in internal [0, 2pi]

    Quote Originally Posted by Chaim View Post
    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.
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  3. #3
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    Re: Find solutions equation in internal [0, 2pi]

    Quote Originally Posted by emakarov View Post
    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!
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  4. #4
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    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?
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