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

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1. May 3, 2017

### Vital

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
cos(2x) = 2 - 5 cos(x)

3. 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!

2. May 3, 2017

### Buffu

Discriminant is wrong. It is 49 you wrote 59.

3. May 3, 2017

### Staff: Mentor

$25 + 24 \neq 59$

Edit: beaten by @Buffu

4. May 3, 2017

### Vital

I am sorry - it's a pure typo. Besides this awkward typo, is this a correct approach?

5. May 3, 2017

### Buffu

Yes I think so, besides that typo you do get $\pi/3$ as the answer.

6. May 3, 2017

### Vital

Thank you very much! )

7. May 3, 2017

### Buffu

You are welcome.

8. May 5, 2017

### epenguin

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.