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

In summary: 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.
  • #1
Vital
108
4

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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  • #2
Discriminant is wrong. It is 49 you wrote 59.
 
  • #3
Vital said:
u = (-5 +- √25 + 24) / 4 = (-5 +- √59) / 4
##25 + 24 \neq 59##

Edit: beaten by @Buffu
 
  • Like
Likes Buffu
  • #4
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?
 
  • #5
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.
 
  • #6
Buffu said:
Yes I think so, besides that typo you do get ##\pi/3## as the answer.
Thank you very much! )
 
  • #7
Vital said:
Thank you very much! )

You are welcome.
 
  • #8
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:
 

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

1. What is an equation?

An equation is a mathematical statement that shows the relationship between two or more variables. It contains an equal sign (=) and is used to solve for unknown values.

2. What are exact solutions?

Exact solutions refer to the specific values of the variables that make the equation true. These values are usually expressed in numerical form and can be verified by substituting them back into the equation.

3. What does [0, 2π) mean in the context of solving an equation?

[0, 2π) is a notation used to indicate a range of values that the solution can take. In this case, it means that the solutions should be within the interval from 0 to 2π, including 0 but not including 2π.

4. How do you solve an equation with exact solutions in [0, 2π)?

To solve an equation in [0, 2π), you need to manipulate the equation using algebraic operations until you isolate the variable on one side. Then, you can plug in values within the given interval and see which one makes the equation true.

5. Why is it important to specify the range of solutions when solving an equation?

Specifying the range of solutions helps to narrow down the possible values and makes it easier to find the exact solutions. It also ensures that the solutions are within a certain range and not just any possible value that makes the equation true.

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