# An interesting one

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

341
Hari Shankar ·

6Ï€?

66
kaymant ·

No.. its not 6Ï€.

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 ·

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

341
Hari Shankar ·

ya 5Ï€ is right.

66
kaymant ·

yes Philip, the answer is indeed 5Ï€.

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 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

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+âˆš7+tan-13-4-âˆš7
From which we have (a+b) as 3Ï€2
i need 2(Ï€+a+b)=5Ï€.

1
Philip Calvert ·

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...

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.

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?