Why isn't the answer the sum of the roots of the given quadratic eqn. i.e, 8/3 ?
Find the sum of all x in [0, 2Ï€] which satisfy the equation
3 cot^{2} x + 8 cot x + 3 = 0

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The question does not ask the sum of the roots of this quadratic (which would be cot x_{1} + cot x_{2}). Rather the question asks to find the sum of those x's which satisfy the give equation in the given range.
The given eqn . 3 cosec^{2}x =  8 cot x
or ,  3 / 4 = sin 2x
but even then ans. is not coming :(
answer is 5Ï€
really sorry for my earlier posts..
its quite simple and straightforward once you put it on paper.
ok
we see both values of cotx satisfying the quadratic will be <0.
also the roots are of the form cot x , 1cot x.
so we can assume the roots to be Î¸ , Ï€+Î¸ and 2Ï€+Î± , Ï€+Î±
such that Î± + Î¸ = Ï€/2 ... (one case when the product of the roots will be =1 )
[ i have taken Î¸ Îµ (Ï€/2,Ï€) and Î± Îµ (Ï€/2 , 0) hence used 2Ï€+Î± instead of Î±]
so sum of all the roots... 4Ï€ + 2(Î±+Î¸) = 5Ï€
it can also be done by taking Î± + Î¸ = 5Ï€2 & use Î± straightaway.. but i think its more or less the same
I did something different.....my method was brute force kind off......
If (a,b) be the roots, then I have a+b=tan^{1}34+âˆš7+tan^{1}34âˆš7
From which we have (a+b) as 3Ï€2
i need 2(Ï€+a+b)=5Ï€.
so it is the same thing na ...?
a+b will be (2n+1)Ï€/2 .. which comes from the observation that
the roots are of the form cot x , 1cot x or as by your method...
afterwards its just addition
Yup it is, philip, let's see what kaymant sir has in his mind, I don't think he'd post something so straightforward under this heading.
I am getting cot(x1+x2) has to be 0. So therefore, to s to be (2n+1)pi/2. Putting n=0,1,2..
shouldn't it be pi/2+3pi/2 only?