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Trig Help PLz

  1. May 21, 2005 #1
    im trying to solve this trig equasion and the book says to factor it i think but its not clear, could anyone help. thxs

    2sin^2 x + 3cos x – 3 = 0
     
  2. jcsd
  3. May 22, 2005 #2
    [tex] sin^2x = 1 - cos^2x [/tex] is all you need. From there its a quadratic.
     
  4. May 22, 2005 #3
    but how do you get the cos^2
     
  5. May 22, 2005 #4

    dextercioby

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    Whozum gave you where that cos^2 comes from.U'd be using the fundamental identity of circular trigonometry

    [tex] \sin^{2}x+\cos^{2}x\equiv 1 \ , \forall \ x\in\mathbb{R}[/tex]

    Daniel.
     
  6. May 22, 2005 #5
    ok heres what i got so far

    [tex] 2sin^2x+3cos x-3 = 0 [/tex]
    [tex] 2(1-cos^2x)+3cos x-3 = 0 [/tex]
    [tex] -2cos^2x+3cos x-1 = 0 [/tex]
    [tex] 2cos^2x-3cos x+1 = 0 [/tex]

    and i know how to factor
    [tex]2x^2-3x+1=0 [/tex]
    [tex](2x-1)(x-1)=0 [/tex]
    [tex]x=1, x=.5 [/tex]

    but how do i solve with the cos
     
  7. May 22, 2005 #6

    dextercioby

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    Well u get the equations

    [tex] a) \ \cos x=1 [/tex]

    [tex] b) \ \cos x=\frac{1}{2} [/tex]

    Fix a domain where u want to solve these equations.

    Daniel.
     
  8. May 22, 2005 #7
    ok, i want to solve it for negative pi to pi
     
  9. May 22, 2005 #8

    dextercioby

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    Well,then i can tell u that the fist equation has only solution if the interval is [itex] \left(-\pi,\pi\right) [/itex],and the second has 2.

    Daniel.
     
  10. May 22, 2005 #9
    i dont get it, do u have to subtract pi from the second equasion
     
  11. May 22, 2005 #10

    dextercioby

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    [tex]\cos x=\frac{1}{2} [/tex] has the solution

    [tex] x=\pm \frac{\pi}{3} [/tex] in the interval u were looking for the solution.

    Daniel.
     
  12. May 22, 2005 #11
    ok i get it the answers are

    for the cos x=1 0
    for the cos x=.5 pi/3 and 5 pi/3
     
    Last edited: May 22, 2005
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