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Analytic Trigonometry

  1. Dec 4, 2006 #1
    1.) sin2x=3cos2
    2.) (3tan2x-1)(tan2x-3)=0
    3.) cos3x=cosx
    4.) 3tan3x=tanx

    Those were a few of the problems in the section I missed on friday due to illness and I was wondering if anyone could walk me through them. Would be greatly appreciated!

    They want to solve the equation for 0<=x<2pie for #1 and #2.
    and
    Find all solutions of the equation in the interval [0,2pie) algebraically for #3 and #4.

    Oh and could anyone tell me how to make the symbols for pie, square root, and greater than or equal to for future reference?
     
  2. jcsd
  3. Dec 4, 2006 #2
    (1) rewrite [tex] \sin^{2}x = 1-\cos^{2} x [/tex]

    (2) set each factor equal to 0


    (3) Let [tex] u = \cos x [/tex]. Then you have [tex] u^{3}-u = 0 [/tex]

    (4) Again let [tex] u = \tan x [/tex]. Then you have [tex] 3u^{3}-u = 0 [/tex]

    The symbols include: \pi, >, < , \leq, \geq, \sqrt{}
     
    Last edited: Dec 4, 2006
  4. Dec 4, 2006 #3
    Thanks a bunch, was a BIG help, I can already see myself loving this website, I have a mediocre teacher and havent been doing the best on tests, even though I ace quizzes, thanks again!
     
  5. Dec 4, 2006 #4

    radou

    User Avatar
    Homework Helper

    Here, regarding the LaTeX code reference: https://www.physicsforums.com/showthread.php?t=8997 (perhaps you already figured it out).
     
  6. Dec 4, 2006 #5
    on 3 and 4 I am still having trouble, I can get 3 to:
    cos3x=cosx
    u3-u=0
    cosx(cos2x-1)=0
    cosx(sin2x)=0

    then get stuck

    and number 4 I cant figure past where you got me
     
  7. Dec 4, 2006 #6
    For (3) [tex] \cos^{2}x-1 \;!= \sin^{2}x [/tex]

    set [tex] \cos x = 0 [/tex] and [tex] \cos^{2}x-1 = 0 [/tex]

    (4) Factor: [tex] u(3u^{2} -1) = 0 [/tex], [tex] u = 0 [/tex] and [tex] 3u^{2} - 1 = 0 [/tex]
     
    Last edited: Dec 4, 2006
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