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The solution to cos x = 2 or any number > 1

  1. Aug 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Generally the inverse cosine of any number > 1

    2. Relevant equations

    cos x = 2

    3. The attempt at a solution

    Obviously by putting this in a calculator, you get an error so the root has to be complex

    I used the identity (e+e-iθ)/2 = cosθ

    through a bit of manipulation, I came to e = 2±√3

    with the solutions being x = ln(2±√3) / i

    is this correct?

  2. jcsd
  3. Aug 25, 2013 #2
    Watch out with your calculations, to keep them clear for yourself and others.
    But, your values for x are indeed correct.
  4. Aug 25, 2013 #3


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    Correct so far but you have not finished. You should be able to get the answer in eityer of the
    "standard forms" a+ bi or [tex]re^{i\theta}[/tex]. First, dividing by i is the same as multiplying by -i so this is [itex]x= -ln(2\pm \sqrt{3})i[/itex]. Next, ln(x) where x is a NEGATIVE real number is itself complex.
  5. Aug 25, 2013 #4


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    True, but 2 - √3 > 0.
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