Solve the equation, exact solutions in [0, 2π)

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Homework Help Overview

The discussion revolves around solving the equation cos(2x) = 2 - 5 cos(x) within the interval [0, 2π). Participants are exploring the algebraic manipulation of the equation and the implications of their findings.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to rearrange the equation and substitute cos(x) with a variable u, leading to a quadratic equation. They express uncertainty about the correctness of their approach and the relevance of a specific solution.
  • Some participants question the calculation of the discriminant and the implications of a typo in the original poster's work.
  • Others suggest that the approach taken by the original poster is generally correct despite the noted typo.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's method and addressing errors in calculations. There is acknowledgment of a potential solution, but no consensus on the overall correctness of the approach has been reached.

Contextual Notes

Participants are navigating through algebraic manipulations and quadratic solutions, with some expressing confusion about the implications of their findings and the specific values derived from the equation.

Vital
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Homework Statement


Hello!

Please, let me know if I am heading towards a correct path in solving the equation. I get stuck in the middle, and obviously head away from the result presented in the book.

Homework Equations


cos(2x) = 2 - 5 cos(x)

The Attempt at a Solution



Gather all on one side:
cos(2x) - 2 + 5 cos(x) = 0

As cos(2x) = 2 (cos(x))2 - 1
2 (cos(x))2 - 1 - 2 + 5 cos(x) = 0
2 (cos(x))2 + 5 cos(x) - 3 = 0

Let cos(x) = u.
Then,
2 u2 + 5 u - 3 = 0

roots of this equation are:
u = (- b +- √b2 - 4ac ) / 2a =>
u = (-5 +- √25 + 24) / 4 = (-5 +- √59) / 4

thus cos(x) = (-5 +- √59) / 4
It is not the end, but I get stuck here because I see that this doesn't seem to lead me to a correct place, because one of the answers is π/3, but if I try to find x from the above expression, I can do it only using

arccos( (-5 +- √59) / 4) = x

Please, let me know if I am doing something wrong.
I have also thought about using Sum to Product formula here, but then I don't get products to work with if I want to set the equation to 0, namely:
as cos(α) + cos(β) = 2cos( (α+β)/2) cos( (α-β)/2)

But I can't use that sum to product formula in cos(2x) - 2 + 5 cos(x) = 0 because the second cosine has a coefficient 5.

Thank you!
 
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Discriminant is wrong. It is 49 you wrote 59.
 
Vital said:
u = (-5 +- √25 + 24) / 4 = (-5 +- √59) / 4
##25 + 24 \neq 59##

Edit: beaten by @Buffu
 
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Buffu said:
Discriminant is wrong. It is 49 you wrote 59.
I am sorry - it's a pure typo. Besides this awkward typo, is this a correct approach?
 
Vital said:
I am sorry - it's a pure typo. Besides this awkward typo, is this a correct approach?

Yes I think so, besides that typo you do get ##\pi/3## as the answer.
 
Buffu said:
Yes I think so, besides that typo you do get ##\pi/3## as the answer.
Thank you very much! )
 
Vital said:
Thank you very much! )

You are welcome.
 
2 u2 + 5 u - 3 = 0 has a fairly simple factorisation hasn't it?

And if you can't see that, you can solve the quadratic the heavy way and knowing the answer when you come to write up pretend that you saw it all along. :oldbiggrin:
 

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