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Trigonometric equation

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    All solutions in the interval 0 ≤ x < 2∏

    Sin 2x = -2cos x

    3. The attempt at a solution

    I went about it in two different ways.

    The first way:

    I simply used the double angle formula to get

    2sinx cosx + 2cos x = 0

    From this we get that whatever makes cos x = 0 will make the whole equation 0.

    There are two answers within the interval; ∏/2 and 3∏/2.

    This is the correct solution.


    When I was studying for my midterm I went about it in a different way though, and I only got one solution. I was wondering what I was doing wrong as it seems to be all legal algebra to me.


    2sinx cosx = -2 cos x

    divide both sides by cos x

    2sinx = -2

    divide by 2

    sin x = -1

    This only gives me one solution, 3∏/2.

    What am I doing wrong?
     
  2. jcsd
  3. Oct 12, 2012 #2

    Mark44

    Staff: Mentor

    This is not the right thing to do. When you do this, you will lose whatever solutions correspond to cos(x) = 0.

    Instead of dividing through by a variable, factor things.

    2sin(x) cos(x) + 2cos(x) = 0

    2cos(x)(sin(x) + 1) = 0

    And so on.
     
  4. Oct 12, 2012 #3
    Ok cool so the first way was correct because when I divide by cos x I'm essentially doing illegal algebra because I technically cannot know if cos x = 0?

    I think I got it. Thank you!
     
  5. Oct 12, 2012 #4

    Mark44

    Staff: Mentor

    Well, you were partly correct in your first way, but you didn't notice that sin(x) = -1 is also a solution. When you factor the expression, both solutions become obvious.
     
  6. Oct 12, 2012 #5
    I actually did do that in my notes. I didn't include it since it doesn't give me an extra solution :).
    Thanks.
     
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