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HELP Trig Equation

  1. Jun 10, 2006 #1
    soo okay...i have this question that im stuck on...

    Solve for x,

    cos 2x - cos^2x = 0, -180degrees<x<180degrees

    i tried solving it...

    cos 2x - cos^2x = 0
    [2cos^2x -1]-cos^2x = 0
    cos^2x -1 = 0 <--(???)
    (cos x -1)(cos x +1) = 0
    therefore... 1)cos x = -1 , 2)cos x = 1

    so...

    1)cos x = -1
    x = cos^-1 (-1)
    x= 180

    2) cos x = 1
    x = 0

    but i have tried on my calculator...and -180 degrees can be one of the answers too...i dont get how to get -180

    SOMEONE HELPP...and reply FASTT plz and Thank yOU in advance
     
  2. jcsd
  3. Jun 10, 2006 #2

    dav2008

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    Gold Member

    Well where is cos(x) equal to -1 besides x=180°?

    (Also, since the original problem asks for solutions greater than -180° and less than 180°, then neither 180° nor -180° would be solutions. Maybe the problem had -180° ≤ x ≤ 180°)
     
    Last edited: Jun 10, 2006
  4. Jun 10, 2006 #3
    the thing is...i dont even know if i did the equation right.... T.T it seems wrong
     
  5. Jun 10, 2006 #4

    dav2008

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    Gold Member

    How does it "seem wrong"? Either you made a mistake and it's wrong or you didn't and it's right.

    I didn't see any mistakes except for the two things I pointed out.

    You can plug in the values of x you found into the original equation to check your work.
     
    Last edited: Jun 10, 2006
  6. Jun 10, 2006 #5
    umm..thanks..
     
  7. Jun 11, 2006 #6

    TD

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    Homework Helper

    You're doing all right, the solution are [itex]k\pi[/itex] with [itex]k \in \mathbb{Z}[/itex].
    The only solution in the given interval is x = 0, if the inequalities weren't strict, you have to add pi and -pi as solutions.
     
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