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Solving a trigonometric equation

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve 2cos(x)^2 + 3cos(x) + 1 = 0 for 0 <= x <= 2pi

    2. Relevant equations
    Trigonometric equations, yadda yadda yadda.

    3. The attempt at a solution
    2cos(x)^2 + 3cos(x) + 1 = 0
    cos(x)(2cos(x) + 3) = -1
    cos(x) = -1/(2cos(x) + 3)

    I then figured out that you get a solution when cos(x) = -1 (so x = pi). However, there are supposed to be multiple solutions, and I don't know how to find these. Any pointers?
     
  2. jcsd
  3. Nov 19, 2012 #2

    SteamKing

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    Notice that the original equation is a quadratic in cos(x).
     
  4. Nov 19, 2012 #3
    Ah, found it. I've been messing around with trigonometric identities for half an hour now. Turns out I just had to factor the whole equation. Thanks. =D
     
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