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Homework Help: How to solve 2 - 2sin^2(theta) = cos(theta) on the domain [0,2pi]

  1. Feb 26, 2005 #1
    [tex] 2-2\sin^2(\theta)=\cos (\theta) [/tex]

    I need to solve this quadratic trig equation algebraically of the domain 0<=x<=2pi I know how to do that but in order to solve the trig functions have to be the same I'm having a little trouble how do I change the cos into sin? It's not cos^2 so we cant use the pythagorean identity.
     
  2. jcsd
  3. Feb 26, 2005 #2

    dextercioby

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    Factor that 2 and use the fundamental identity
    [tex] \sin^{2}x+\cos^{2}x =1 [/tex]

    Daniel.
     
  4. Feb 26, 2005 #3
    Oh thanks heres what I got

    [tex] 2(1-\sin^2\theta)=\cos \theta [/tex]
    [tex] 2(\cos^2\theta)=\cos\theta [/tex]

    Expand [tex] 2\cos^2\theta=\cos\theta [/tex]

    Rearrange [tex] 2\cos^2-\cos\theta=0 [/tex]
     
  5. Feb 26, 2005 #4

    dextercioby

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    Factor cos\theta and end up with 2 simple equations...

    Daniel.
     
  6. Feb 26, 2005 #5
    [tex] \cos \theta=0[/tex] and [tex] \cos\theta=1/2 [/tex]
     
  7. Feb 26, 2005 #6

    dextercioby

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    Perfect.Sove each of them and then write the final solution to the initial eq.

    Daniel.
     
  8. Feb 26, 2005 #7
    I got 300 degrees and 60 degrees as solutions for cos (x)= 1/2

    but Im not sure for cos(x)=0

    its either 0 degrees and 180 degrees, or 90 degrees
     
  9. Feb 26, 2005 #8

    BobG

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    Try 90 and 270.

    Cos (0) is 1
    Cos (180) is -1
     
  10. Feb 26, 2005 #9
    ok so are there 4 solutions

    90 degrees 270 degrees, 300 degrees and 60 degrees

    Is that correct?
     
  11. Feb 27, 2005 #10

    dextercioby

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    Yes,u can make a graph (trigon.circle) to check the validity.


    Daniel.
     
  12. Feb 27, 2005 #11
    but my answers are right? :rofl:
     
  13. Feb 27, 2005 #12

    dextercioby

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    Yes,they are...And what's so funny...?My new haircut...?:confused:

    Daniel.
     
  14. Feb 27, 2005 #13
    lol thanks sooooooooooooo much i finally got something thats the funny part! :tongue2:
     
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