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Homework Help: How would I solve sin2Ѳ+sin3Ѳ=0?

  1. Sep 1, 2014 #1
    EDIT: Please delete...was very confused and have now found solution!
     
    Last edited: Sep 1, 2014
  2. jcsd
  3. Sep 2, 2014 #2

    Simon Bridge

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    How about posting the answer instead, then someone else confused can benefit from your confusion ;)
     
  4. Sep 2, 2014 #3
    Sure!
    Use sin(A+B)/2*cos(A-B)/2
    =2sin (3x+2x)/2*cos (2x-3x)/2
    2sin(5x)=0
    Thus sinx=0
    x=0, 180+2 (pi)n
    2cos (-x)/2
    Thus cosx=1/2
    x=90+2 (pi)n
     
  5. Sep 2, 2014 #4

    ehild

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    Do you mean that Θ=pi/2 is a solution?
    But sin(2Θ)+sin(3Θ)=sin(180°)+sin(270°)=-1.

    And what about Θ=72°? Is not it a solution, too?

    ehild
     
  6. Sep 2, 2014 #5
    Well, the solutions manual is getting:
    Θ=0, 2∏/5, 4∏/5, ∏, 6∏/5, 8∏/5
    The interval is also 0 ≤ θ ≤2∏

    I forgot to include the interval before, but could figure it out without the interval. Now I don't know where to go with the problem as it looks like I'm solving it completely wrong.
     
    Last edited: Sep 2, 2014
  7. Sep 2, 2014 #6

    ehild

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    No, you started very well, it was a splendid idea.
    But you miss some parentheses. It should be 2sin [(3x+2x)/2]*cos [(2x-3x)/2]
    So sin(5x/2)=0, and from this does not follow that sinx=0. Instead, sin(5x/2)=0 that is, 5x/2=kπ. x=???

    Do the same with the other factor, cos(x/2).

    ehild
     
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