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Find the locations where the slope = 0

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    I just had a test I can't remember exactly how the question went but it was a relatively easy differentiation problem. I can only remember my result, which was

    y' = 2sinxcosx - sinx


    2. Relevant equations



    3. The attempt at a solution

    I noticed that 2sinxcosx is also sin2x so you could rewrite it as Sin2x - Sinx

    Now, to find the places where the slope is 0 i obviously just set it to 0 and solve. My problem was that I didn't know if I should go the Sin2x - Sinx route or Factor out a sine and do it like thisL

    sinx(2cosx -1) = 0

    cosx = 1/2

    x = ∏/3 +2∏k

    x = 5∏/3 + 2∏k

    where k is any integer


    The above one is the route I chose to take at the end. However, if I solve for sin2x - sin x = 0

    I can just say, ∏k, where k is any integer? Where is my logic flawed?
     
  2. jcsd
  3. Apr 3, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    x=pi*k aren't the only solutions to sin(2x)-sin(x)=0. They are just some of them. You've found the others in your first analysis. If sin(x)*(2*cos(x)-1)=0 then either sin(x)=0 or 2*cos(x)-1=0.
     
  4. Apr 3, 2013 #3
    Thanks for the answer though.
     
    Last edited by a moderator: Apr 4, 2013
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