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

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The discussion revolves around solving the equation cos(2x) = 2 - 5 cos(x) and the user's attempt to derive the correct solutions. The user correctly rearranges the equation and substitutes cos(x) with u, leading to the quadratic equation 2u² + 5u - 3 = 0. A typo in the discriminant calculation is pointed out, correcting it to 49 instead of 59, which affects the roots. Despite the confusion, the approach is validated, and the user is reassured that they can indeed arrive at the solution x = π/3. Overall, the discussion emphasizes the importance of careful calculations and confirms that the method used is fundamentally sound.
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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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