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    Really stuck on this qu. any help would be much appreciated

    given that f(x) = cot x + 3

    solve f(x) + f '(x) = 0, for 0 < x < 2 pie

    i dont have a clue. i tried adding cot x + 3 to -cosec^2x and tried to make every in cos but failed. what should I do?

    thanks. rep on offer
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    (Original post by Tommy Jay)
    Really stuck on this qu. any help would be much appreciated

    given that f(x) = cot x + 3

    solve f(x) + f '(x) = 0, for 0 < x < 2 pie

    i dont have a clue. i tried adding cot x + 3 to -cosec^2x and tried to make every in cos but failed. what should I do?

    thanks. rep on offer
    Okay so we have cot(x)+3-cosec^2(x)=0. cos(x)/sin(x)+3-1/sin^2(x)=0. => cos(x)sin(x)+3sin^2(x)-1=0 => cos(x)sin(x)=1-3sin^2(x); at this point I'd be tempted to double both sides then put it all in one trig function type (either sin(x) or cos(x))
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    (Original post by Tommy Jay)
    Really stuck on this qu. any help would be much appreciated

    given that f(x) = cot x + 3

    solve f(x) + f '(x) = 0, for 0 < x < 2 pie

    i dont have a clue. i tried adding cot x + 3 to -cosec^2x and tried to make every in cos but failed. what should I do?

    thanks. rep on offer
    As cosec^2(x) = 1 + cot^2 (x) , you can substitute.

    So f(x) + f '(x) = cot(x) + 3 - (1 + cot^2(x)) = 0

    So, -1 - cot^2(x) + 3 + cot(x) = 0

    Therefore, cot^2 (x) - cot(x) -2 =0
    Then you factorise which gives you :

    = [cot(x) + 1 ] [cot(x)-2] = 0

    so cot(x) = -1 , cot(x) = 2

    Then you should be able to work it out from there :yes:
 
 
 
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