66No.. its not 6Ï€.
Find the sum of all x in [0, 2Ï€] which satisfy the equation
3 cot2 x + 8 cot x + 3 = 0
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13 Answers
1Why isn't the answer the sum of the roots of the given quadratic eqn. i.e, 8/3 ?
66The question does not ask the sum of the roots of this quadratic (which would be cot x1 + cot x2). Rather the question asks to find the sum of those x's which satisfy the give equation in the given range.
1The given eqn . 3 cosec2x = - 8 cot x
or , - 3 / 4 = sin 2x
but even then ans. is not coming :(
1answer is 5Ï€
really sorry for my earlier posts..
its quite simple and straightforward once you put it on paper.
1ok
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 θ , π+θ and 2π+α , π+α
such that α + θ = π/2 ... (one case when the product of the roots will be =1 )
[ i have taken θ ε (π/2,π) and α ε (-π/2 , 0) hence used 2π+α instead of α]
so sum of all the roots... 4π + 2(α+θ) = 5π
it can also be done by taking α + θ = 5π2 & use α straightaway.. but i think its more or less the same
11I did something different.....my method was brute force kind off......
If (a,b) be the roots, then I have a+b=tan-13-4+√7+tan-13-4-√7
From which we have (a+b) as 3Ï€2
i need 2(Ï€+a+b)=5Ï€.
1so 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
11Yup 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?