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Simplify and Solve Trig Equation

  1. Jan 14, 2007 #1

    Gib Z

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

    Simplify and Solve the Following Trigonometric Equation

    2. Relevant equations

    [tex]\cos^2 x - 3\cos x - 2\sin x +2 =0 [/tex]

    3. The attempt at a solution

    I've changed the expression, but it doesn't seem any better...

    I've got [tex]\frac{\cot x \cdot (\cos (x) -3)}{2}=1 - \csc x[/tex]

    But that doesn't seem to help me at all...Stuck badly.
     
  2. jcsd
  3. Jan 14, 2007 #2
    Solving equations that only contain sin x or cos x is easier than an equation that contains both.
     
  4. Jan 14, 2007 #3

    Gib Z

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    Yes, I know how to solve those ones. Unfortunately this one didn't come this way...
     
  5. Jan 14, 2007 #4

    mjsd

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    try using the fact that if [tex]t=\tan\frac{x}{2}[/tex] then
    [tex]\displaystyle{\cos x = \frac{1-t^2}{1+t^2}}[/tex] and
    [tex]\displaystyle{\sin x = \frac{2t}{1+t^2}}[/tex], solve for [tex]t[/tex], should be a quatic eqn and then you can solve for [tex]x[/tex].

    NB: may not be the simplest way to do things, but at least in principle this will work. would like to see if there is a simpler way to do this....
     
  6. Jan 14, 2007 #5

    robphy

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    Upon substituting [tex]\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix}})[/tex] and [tex]\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix}})[/tex], where [tex]i^2=-1[/tex], you end up with a quartic polynomial in one variable "[tex]e^{ix}[/tex]", which generally has four solutions.

    If the above method is not allowed or not what is expected, you probably have to rewrite the equation in terms of one trigonometric-function of x [using trig identities], as suggested by Moridin.
     
  7. Jan 14, 2007 #6

    Gib Z

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    I have absolutely no idea how to do this using identities, just makes anything I do more messy and doesnt help. And I don't think this problem should require math from that level, but ill try it. Solving quartics are hard though...
     
  8. Jan 14, 2007 #7

    robphy

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    I (hopefully correctly) typed in the equation to Maple and asked for solutions. Two of the four roots for x are real (and will probably arise simply).. the other two are complex and are not pretty (though trig function of these may be prettier).
     
  9. Jan 14, 2007 #8

    mjsd

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    if you use my method, the quartic in [tex]t[/tex] that you need to solve is actually very easy to do! try it! Hint: there are two real integer roots, the remaining complex roots can then be easily worked out using the quadratic equation. Finding [tex]x[/tex] is then a matter of inverting the [tex]\tan[/tex].
    NB: robphy's method is in essence the same; the only difference is that you introduce complex numbers from the very beginning.
     
  10. Jan 14, 2007 #9

    Gib Z

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    Maybe there is a simpler way. I didn't actaully post my original question, but what I got it to correctly, heres the whole thing.

    cos^3 x-3cos^2 x+ cos x = 2cos(x/2 + pi/4)sin(3x/2 - pi/4)
    cos^3 x-3cos^2 x+ cos x= sin2x - cos x
    cos^3 x-3cos^2 x+ 2cos x - 2sinxcosx=0
    cos x(cos^2 - 3cos x +2 - 2sin x)=0

    Now I'll do the first one, cos x=0, now I needed the 2nd part. Was there an easier way from the start?

    Ill try using the t, looks good.
     
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