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    Really stuck on c3 question. any help appreciated:

    given that f(x) = cotx + 3

    solve f(x) + f '(x) = 0 for 0 < x < 2pie, to 1 d.p

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    A good place to start might be to calculate f'(x)...
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    (Original post by Kolya)
    A good place to start might be to calculate f'(x)...
    -cosec^2 x. but then i get stuck...
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    (Original post by Kolya)
    A good place to start might be to calculate f'(x)...
    if you could help me you would save my life. if i get this qu. done i can go to bed lol
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    If I tell you that 1 + cot^2(x) = cosec^2(x) that should help
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    well  cotx=\frac{cosx}{sinx} and  cosec^2x=\frac{1}{sin^2x}

    plus  1+cot^2x=cosec^2x as said above

    re arrange it from there and you should be on the right track
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    Actually no, all you need to know is what DLS said! I gave you too much info!
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    (Original post by DLS)
    If I tell you that 1 + cot^2(x) = cosec^2(x) that should help
    ahhh cheers . i knew i was missing something. i am now left with a quadratic which factorises. Bed here i come woooo
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    (Original post by Tommy Jay)
    Really stuck on c3 question. any help appreciated:

    given that f(x) = cotx + 3

    solve f(x) + f '(x) = 0 for 0 < x < 2pie, to 1 d.p

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    Cotx +3 -Cot^2 x -1 =0
    Cot^2 x - Cot x -2 = 0
    (C-2)(C+1)
    cotx = 2
    cotx=-1
    tanx=(1/2)
    tanx=(-1)
    etc.
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    What exam board was this ?
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    (Original post by Tommy Jay)
    ahhh cheers . i knew i was missing something. i am now left with a quadratic which factorises. Bed here i come woooo
    LOL.
    Good luck!
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    (Original post by n_251)
    well  cotx=\frac{cosx}{sinx} and  cosec^2x=\frac{1}{sin^2x}

    plus  1+cot^2x=cosec^2x as said above

    re arrange it from there and you should be on the right track
    How are we supposeed to know the differential of Cotx ?
    y = 1/tanx

    can I use the quotient rule to find ... =cosec^2 x?
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    I think it's edexcel ?
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    (Original post by Maturity)
    How are we supposeed to know the differential of Cotx ?
    y = 1/tanx

    can I use the quotient rule to find ... =cosec^2 x?
    If you're talking about edexcel you'll not be expected to learn it. It will be given to you in the formula booklet.
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    I think its edexcel too, all given in the formula booklet except sinx and cosx
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    (Original post by FZka)
    If you're talking about edexcel you'll not be expected to learn it. It will be given to you in the formula booklet.
    AQA, no I looked for it and it is not in the given formulae to learn.
    But I was wondering how would you actually go about solving it? Anyways.
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    (Original post by Maturity)
    How are we supposeed to know the differential of Cotx ?
    y = 1/tanx

    can I use the quotient rule to find ... =cosec^2 x?
    Here you go. Check C3.
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  1. File Type: pdf maths-GCE-formula-book.pdf (548.1 KB, 41 views)
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    (Original post by Maturity)
    AQA, no I looked for it and it is not in the given formulae to learn.
    But I was wondering how would you actually go about solving it? Anyways.

    ohh just saw the message. after i uploaded the formula book.
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    f(x) = cot(x) + 3

    First find f'(x)

    f'(x) = (d/dx) (cox)(sinx)^-1
    f'(x) = -sin(x)(sin(x))^-1 - (sinx)^-2(cosx)^2
    f'(x) = -1 - cot^2(x)

    then f(x) + f'(x) = 0 is the same as...

    cot^2(x) - cot(x) - 2 = 0

    then you just have a quadratic.

    (cotx - 2)(cotx + 1) = 0

    cot(x) = 2 or cot(x) - 1

    x = arctan(0.5) or x = arctan(-1)

    then you can find all solutions in the given range.

    Hope this helps.
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    (Original post by FZka)
    ohh just saw the message. after i uploaded the formula book.
    Thank you anyway .
 
 
 
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