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Stuck on this hard trig problem

  1. Aug 21, 2004 #1
    ok heres the problem i haven't taken trig for like 2 years and suddenly i have been thrown back in to it...

    problem: 2 cos 2x - 4 cos x + 3 = 0 0 < x < 2 pi

    my work (used cos 2 x = 2 cos^2 x - 1)

    4 cos^2 x - 2 - 4 cos x + 3 = 0

    4 cos^2 x - 4 cos x + 1 = 0

    ok now here is my problem if i try and factor it out i ll be left with the a really ugly 1/ ( 4 cos x)

    is there another equation i could use... is my logic flawed... is there an easier way... do i just have to do it the ugly way... ???
     
  2. jcsd
  3. Aug 21, 2004 #2
    2cos2x - 4cosx + 3 = 0
    2(2cos^{2}x-1) - 4cosx + 3 = 0
    4cos^{2}x - 4cosx + 1 = 0
    (2cosx-1)^{2} = 0
    cosx = 1\2
    x = 60 deg. or 300 deg.
     
  4. Aug 21, 2004 #3
    ... me stupid...

    PS having a little to much fun with pi arn't we... you should go out to atleast when pi gets to 123456789......
     
  5. Aug 22, 2004 #4

    HallsofIvy

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    "There is not enough room in this margin..." :wink:
     
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