# The quadratic trigonometric equation cot^2(x)-b*cot(x)+c=0 has solutions pi/6, pi/7, 7pi/6, 5pi/4 in the interval [0, 2pi]. What are the values of b and c?

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### 1 Answer

This is a quadradic in cot(x)

We can use u = cot(x)

u^2 - bu + c = 0

solution is `u = (b+-sqrt(b^2-4c))/(2)`

I think you meant `pi/4` not `pi/7` it doesn't make sense if that is not true

`cot(pi/6) = cot((7pi)/6) = sqrt(3)` and

`cot(pi/4) = cot(5pi/4) = 1`

So `(b+-sqrt(b^2-4c))/(2) = sqrt(3) or 1`

`(b-sqrt(b^2-4c))/2 = 1`

`b-sqrt(b^2-4c) = 2`

`sqrt(b^2 - 4c) = b-2`

`b^2 - 4c = b^2-4b+4`

`-4c = -4b+4`

`c = b - 1`

and

` (b+sqrt(b^2-4c))/2 = sqrt(3)`

`b+sqrt(b^2-4c) = 2sqrt(3)`

`sqrt(b^2-4c) = 2sqrt(3) - b`

`b^2 - 4c = 2^2sqrt(3)^2 - 4sqrt(3)b + b^2`

`-4c = 12 - 4sqrt(3)b`

`c = sqrt(3)b - 3`

Solving for b we get

`sqrt(3)b - 3 = b - 1`

`(sqrt(3)-1)b = 2`

`b = 2/(sqrt(3)-1)*(sqrt(3)+1)/(sqrt(3)+1) = 2(sqrt(3) + 1)/(3 - 1)=sqrt(3)+1`

`c = b - 1 = sqrt(3)`

We can check:

`cot(x)^2 - (sqrt(3)+1)cot(x) + sqrt(3) = 0`

If `x = pi/4 or 5pi/4 cot(x) = 1`

`1^2 - (sqrt(3)+1)(1) + sqrt(3) = 1 - sqrt(3) - 1 + sqrt(3) = 0` which checks

if `x = pi/6 or 7pi/6 cot(x) = sqrt(3)`

`(sqrt(3))^2 - (sqrt(3)+1)(sqrt(3) + sqrt(3) = 3 - (sqrt(3))(sqrt(3)) - sqrt(3) + sqrt(3) = 3 - 3 - sqrt(3) + sqrt(3) = 0` which checks

So the answers are

`b = sqrt(3)+1` and `c = sqrt(3)`