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Trig double-angle itdentities

  1. May 18, 2006 #1
    ok heres a question i cant figure out why the answer is the way it is.

    Solve each equation for all values of x.

    1.cos2x=cosx

    *First i use the double identity 2cos²x-1 for cos2x and got:
    2cos²x-1=cosx

    *then i subtracted cosx and got:
    2cos²x-cosx-1=0

    *then i subtract 1 on both side and got:
    2cos²x-cosx=1

    *I then factor 2cos²x-cosx and got:
    cosx(2cosx-1)=1

    *then i set them equal to 1 and got this:

    cosx=1
    =0°

    2cosx-1=1
    added one and divided by 2 to get:
    cosx=1
    =0°


    Since it is cosine i set it like this: 0°+360k°

    But my book says the correct answer is 0°+120k°

    Please someone help me with this.
    And i may have more questions on some other double identity problems.
     
  2. jcsd
  3. May 18, 2006 #2

    George Jones

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    Why do both factors have to equal 1? For example, (-3)*(-1/3) = 1. Note, I'm not saying these numbers are useful for your question, I'm just pointing out other possibilities.

    Try factoring your equation before moving 1 to the RHS.

    Regards,
    George
     
  4. May 18, 2006 #3

    VietDao29

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    So far so good. From here, it's just the quadratic equation in cos x, i.e if you substitute t = cos x, your equation will become:
    2t2 - t + 1 = 0.
    Can you solve for cos x (or t?) from the equation? From there, I think you'll be able to solve for x. Can you go from here? :)
     
  5. May 18, 2006 #4
    So that means i mest up somewhere in the factoring? and its not suppose to both equal 1?

    Well, i get this after i factor 2cos²x-cosx-1=0 :

    cosx(2cos-1)-1=0

    I am really stuck on this problem.
     
  6. May 18, 2006 #5

    George Jones

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    Factor the trinomial into a product of binomials.

    Big hint: substitute y = cosx.

    Regards,
    George
     
  7. May 18, 2006 #6
    so it would be: 2y^2-y+1=0

    and then would the product of the binomials would be (2y+1)(y-1)?
     
  8. May 18, 2006 #7

    George Jones

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    Right, and the product equals zero. What does this tell you?

    Regards,
    George
     
  9. May 18, 2006 #8
    Oh yeah!! i see now!

    it will come out to cosx=-1/2 and 1 which is "120°" and "0°" which is why it has to be 0°+120k°.

    It was just some simple wrong factoring that got me on the wrong track.
    Thanks!
     
  10. May 18, 2006 #9

    George Jones

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    Good!

    Sorry VietDao29 - I didn't see your post.

    Regards,
    George
     
  11. May 18, 2006 #10

    0rthodontist

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    Just to illustrate why the terms of your earlier product don't have to equal 1: (1/3)*3 = 1 but neither is 1.
     
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