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Solving trigomonetry equation for x

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve
    [tex]sin^3 x + cos^3 x + \frac{1}{4}(sin x - cos x) = \frac{cos 2x}{cos x - sin x}[/tex]


    2. Relevant equations
    trigonometry


    3. The attempt at a solution
    After putting some effort, I got: sin 4x - sin 2x + 1 = 0

    I don't know how to proceed....

    Thanks
     
  2. jcsd
  3. Dec 4, 2011 #2

    gb7nash

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    Set u = sin2x. What's your new equation?
     
  4. Dec 4, 2011 #3
    I don't get your hint.

    sin 4x - sin 2x + 1 = 0
    2 sin 2x cos 2x - 2 sin x cos x + 1 = 0
    4 sin x cos x (1 - 2 sin2x) - 2 sin x cos x + 1 = 0
    4 sin x cos x (1 - 2u) - 2 sin x cos x + 1 = 0

    and then...:confused:

    Should I change u = sin2x to sin x = √u then draw triangle to find cos x in term of u? I think it will be more complicated
     
  5. Dec 4, 2011 #4

    gb7nash

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    You're thinking into this way too much.

    Starting from sin4x - sin2x + 1 = 0, make a simple substitution u = sin2x and plug the u stuff into the equation.
     
  6. Dec 4, 2011 #5
    How can you get that equation?
     
  7. Dec 4, 2011 #6

    gb7nash

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    Edit:

    My mistake, I thought you meant sin4x, not sin (4x). Ignore my last post!
     
  8. Dec 4, 2011 #7
    So, do you have new idea? :smile:

    Or maybe it is not solvable...
     
  9. Dec 4, 2011 #8

    Mentallic

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    You've made a mistake somewhere because the original equation and your final equation aren't equivalent.
     
  10. Dec 4, 2011 #9
    [tex]sin^3 x + cos^3 x + \frac{1}{4}(sin x - cos x) = \frac{cos 2x}{cos x - sin x}[/tex]

    [tex](sin x + cos x) (sin^2 x - sin x cos x + cos^2 x) - \frac{1}{4}(cos x - sin x) = \frac{cos 2x}{cos x - sin x}[/tex]

    [tex]cos 2x (1 - sin x cos x) - \frac{1}{4}(1 - sin 2x) = cos 2x[/tex]

    [tex]4 cos 2x sin x cos x + 1 - sin 2x = 0[/tex]

    [tex]sin 4x - sin 2x + 1 = 0[/tex]

    Correct?
     
  11. Dec 4, 2011 #10

    Mentallic

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    Sorry, that's my mistake... The expressions aren't equivalent but the solution sets are :biggrin:
    minus the [itex]x\neq n\pi+\pi/4[/itex] of course.

    It doesn't seem as though this is simple to solve. Were you expecting a numerical solution?
     
  12. Dec 4, 2011 #11
    I don't think so; we haven't covered numerical solution.
     
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