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Solve the equation, exact solutions in [0, 2π)

  1. May 3, 2017 #1
    1. The problem statement, all variables and given/known data
    Hello!

    Please, let me know if I am heading towards a correct path in solving the equation. I get stuck in the middle, and obviously head away from the result presented in the book.

    2. Relevant equations
    cos(2x) = 2 - 5 cos(x)

    3. The attempt at a solution

    Gather all on one side:
    cos(2x) - 2 + 5 cos(x) = 0

    As cos(2x) = 2 (cos(x))2 - 1
    2 (cos(x))2 - 1 - 2 + 5 cos(x) = 0
    2 (cos(x))2 + 5 cos(x) - 3 = 0

    Let cos(x) = u.
    Then,
    2 u2 + 5 u - 3 = 0

    roots of this equation are:
    u = (- b +- √b2 - 4ac ) / 2a =>
    u = (-5 +- √25 + 24) / 4 = (-5 +- √59) / 4

    thus cos(x) = (-5 +- √59) / 4
    It is not the end, but I get stuck here because I see that this doesn't seem to lead me to a correct place, because one of the answers is π/3, but if I try to find x from the above expression, I can do it only using

    arccos( (-5 +- √59) / 4) = x

    Please, let me know if I am doing something wrong.
    I have also thought about using Sum to Product formula here, but then I don't get products to work with if I want to set the equation to 0, namely:
    as cos(α) + cos(β) = 2cos( (α+β)/2) cos( (α-β)/2)

    But I can't use that sum to product formula in cos(2x) - 2 + 5 cos(x) = 0 because the second cosine has a coefficient 5.

    Thank you!
     
  2. jcsd
  3. May 3, 2017 #2
    Discriminant is wrong. It is 49 you wrote 59.
     
  4. May 3, 2017 #3

    DrClaude

    User Avatar

    Staff: Mentor

    ##25 + 24 \neq 59##

    Edit: beaten by @Buffu
     
  5. May 3, 2017 #4
    I am sorry - it's a pure typo. Besides this awkward typo, is this a correct approach?
     
  6. May 3, 2017 #5
    Yes I think so, besides that typo you do get ##\pi/3## as the answer.
     
  7. May 3, 2017 #6
    Thank you very much! )
     
  8. May 3, 2017 #7
    You are welcome.
     
  9. May 5, 2017 #8

    epenguin

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    Homework Helper
    Gold Member

    2 u2 + 5 u - 3 = 0 has a fairly simple factorisation hasn't it?

    And if you can't see that, you can solve the quadratic the heavy way and knowing the answer when you come to write up pretend that you saw it all along. :oldbiggrin:
     
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