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# C3 Help watch

1. 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
2. (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))
3. (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

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