An interesting one

Find the sum of all x in [0, 2π] which satisfy the equation
3 cot2 x + 8 cot x + 3 = 0

13 Answers

341
Hari Shankar ·

ya 5π is right.

66
kaymant ·

yes Philip, the answer is indeed 5π.
What about the solution?

1
Philip Calvert ·

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 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 = 2 & use α straightaway.. but i think its more or less the same

1
Shashank Holla ·

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?

11
Devil ·

I 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 2
i need 2(π+a+b)=5π.

1
Philip Calvert ·

so it is the same thing na ...?

a+b will be (2n+1)π/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

11
Devil ·

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.

341
Hari Shankar ·

6π?

66
kaymant ·

No.. its not 6π.

1
Maths Musing ·

Why isn't the answer the sum of the roots of the given quadratic eqn. i.e, 8/3 ?

66
kaymant ·

The 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.

1
Maths Musing ·

The given eqn . 3 cosec2x = - 8 cot x
or , - 3 / 4 = sin 2x
but even then ans. is not coming :(

1
Philip Calvert ·

answer is 5π

really sorry for my earlier posts..
its quite simple and straightforward once you put it on paper.

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