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Solving trig equations

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve sin2θ - 1 = cos2θ in range 0 ≤ θ ≤ 360°

    2. Relevant equations



    3. The attempt at a solution

    I always struggle with the end of these questions, deciding which answers are correct.

    Here's what I have done;

    let 2θ = x

    cosx + 1 = sinx

    cos2x + sin2x = 1
    sin x = √(1 - cos2x)

    cosx + 1 = √(1 - cos2x)

    square both sides

    cos2x + 2cosx + 1 = 1 - cos2x

    2cos2x + 2cosx = 0

    factor out 2cosx

    2cosx (cosx + 1) = 0

    cosx = 0 when x = 270, 90

    so θ = 135, 45

    cosx - 1 = 0

    x = 180, 270

    θ = 90, 135

    Only 90 and 45 are in the first quadrant where cos is +ve, does that mean these are the answers I want?

    Thanks for any help you can give, I hope my question is clear.
     
  2. jcsd
  3. Nov 26, 2013 #2

    adjacent

    User Avatar
    Gold Member

    What is cos2 and sin2? It does not have any value.
     
  4. Nov 26, 2013 #3
    it's cos2 of x, which i'm using to represent 2θ

    I don't really understand what you're asking
     
  5. Nov 26, 2013 #4
    Recheck this. What are the solutions to cos(x)=1?
     
  6. Nov 26, 2013 #5
    cosx = 1 when x = 0 or 360.
     
  7. Nov 26, 2013 #6

    Mark44

    Staff: Mentor

    What's the reason for your comment, adjacent? I don't see anything wrong with the notation that BOAS used.
     
  8. Nov 26, 2013 #7
    Since ##0 \leq \theta \leq 2\pi##, therefore ##0\leq x \leq 4\pi##, you should find solutions for x in this range.
     
  9. Nov 26, 2013 #8

    adjacent

    User Avatar
    Gold Member

    It's because cos itself does not have any value,it's a function.Then how can cos2 have any value?

    For example, ##f(x)=2x+1##
    ##f## itself has no value nor ##f^2##


    I know I am wrong somewhere and I would appreciate if you clarify my doubt.
     
  10. Nov 26, 2013 #9
    this does make sense, but I struggle with the logic of which solutions are the ones I want.

    In my OP at the end, I justified the solutions i 'chose' by saying that they were the only ones in a quadrant where cos is +ve.

    Is this what I want?

    i.e in the range ##0\leq x \leq 4\pi## there are two revolutions to go through, so I think there should be another set of solutions that lie within the first quadrant.
     
  11. Nov 26, 2013 #10
    It can be positive in both the first and fourth quadrants.
     
  12. Nov 26, 2013 #11
    No, i'm somewhat confused.

    cos(x) = 0

    x = 90, 270, 450, 630

    so θ = 45, 135, 225, 315

    cos(x) = 1

    x = 0, 360, 720 (I think i'm correct not to include the -1 values here)

    so θ = 0, 180, 360
     
  13. Nov 26, 2013 #12
    Looks very good to me but it isn't necessary that the values of you have found are the solutions to the original equation.

    While squaring, there is always a risk of getting extra solutions so the best method is to find a way which does not involve squaring.

    For the given case, you can use:

    sin(2θ)=2sinθcosθ and cos(2θ)=2cos2θ-1

    Are you aware of the above identities?
     
  14. Nov 26, 2013 #13
    They're not committed to memory, but they look familiar.

    I shall rework the question using them and see how I get on. Thanks for helping me.
     
  15. Nov 26, 2013 #14

    Mark44

    Staff: Mentor

    Are you asking about this? cos2x + sin2x = 1

    BOAS write several equations that involved cos2x or sin2x, and these are well-understood forms of notation.

    cos2x means exactly the same thing as (cos(x))2.
     
  16. Nov 26, 2013 #15
    If, in your first equation, you move the cosx to the right side of the equation, and square the resulting relationship, you get

    cos2x+sin2x-2sinxcosx=1
    So, sinxcosx=0. So, at the solution points, either sinx is zero or cosx is zero. You have to check with the original equation to see which such points are allowed.
     
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