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    i differentiated f(x) and got
    f'(x)= -2sin2x+cosx
    =-4sinxcosx+cosx
    i made f'(x)=0
    0=-4sinxcosx+cosx
    4sinxcosx=cosx
    4sinx=1
    sinx=1/4
    x=2.89 radians
    but the answers also include 0.25, 4.71,1.57 radians
    can someone explain to me why these are included, please include full working out as well.
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    (Original post by assassinbunny123)
    i differentiated f(x) and got
    f'(x)= -2sin2x+cosx
    =-4sinxcosx+cosx
    i made f'(x)=0
    0=-4sinxcosx+cosx
    4sinxcosx=cosx
    4sinx=1
    sinx=1/4
    x=2.89 radians
    but the answers also include 0.25, 4.71,1.57 radians
    can someone explain to me why these are included, please include full working out as well.
    are you sure  f'(x) = -4sin(x)cos(x) + cos(x) ?
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    (Original post by assassinbunny123)
    i differentiated f(x) and got
    f'(x)= -2sin2x+cosx
    =-4sinxcosx+cosx
    i made f'(x)=0
    0=-4sinxcosx+cosx
    4sinxcosx=cosx
    4sinx=1
    sinx=1/4
    x=2.89 radians
    but the answers also include 0.25, 4.71,1.57 radians
    can someone explain to me why these are included, please include full working out as well.
    Rather than dividing by \cos x you should factor it out! \cos x = 0 provides the leftover solutions
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    (Original post by assassinbunny123)
    i differentiated f(x) and got
    f'(x)= -2sin2x+cosx
    =-4sinxcosx+cosx
    i made f'(x)=0
    0=-4sinxcosx+cosx
    4sinxcosx=cosx
    4sinx=1
    sinx=1/4
    x=2.89 radians
    but the answers also include 0.25, 4.71,1.57 radians
    can someone explain to me why these are included, please include full working out as well.
    0.25 is another solution to sin(x) = 1/4 and the other two solutions come from considering the case that cos(x) = 0.
    When you divide both sides by cos(x) you lose these solutions so be careful of that, it's also important to bear in mind that sin or cos will have 2 points at which they = any given value per cycle of 2 radians
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    (Original post by BTAnonymous)
    are you sure  f'(x) = -4sin(x)cos(x) + cos(x) ?
    yes i used the double angle formula sin2x=2sinxcosx hence -2sin2x=-4sinxcosx.
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    (Original post by RDKGames)
    Rather than dividing by \cos x you should factor it out! \cos x = 0 provides the leftover solutions
    should i always factories rather than dividing? when i have both sinx and cosx.
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    (Original post by assassinbunny123)
    should i always factories rather than dividing? when i have both sinx and cosx.
    Yes. You may only divide when you know for certain that the expression you're dividing by cannot equal 0
 
 
 
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