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Little Help With a Math Question

  1. Dec 31, 2004 #1
    Can't figure out this question, a little help would be great.

    Solve for X, domain: 0<x<2pi

    tan4X - tan2X = 0

    What i got is that if we use the double angle formula to expand tan2X into

    2tanx / 1-tan^2x

    the u move it onto the other side and then somehow tan2x is = to tan4X, i'm very confused, please help.
     
  2. jcsd
  3. Dec 31, 2004 #2
    for tan 4x you can set it up as tan 2(2x) = 2 tan2x/(1-(tan 2x)^2). Don't simplify tan 2x and just substitute the simplifed form of tan 4x into the equation. The equation is already set to 0 so just solve for x.

    Edit: don't forget about the non-permissable values..
     
    Last edited: Dec 31, 2004
  4. Dec 31, 2004 #3

    dextercioby

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    [tex] \tan 4x=\tan 2x [/tex] (1)

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

    Plug (2) in (1) and solve the eq.Be careful with the 5 points 0,pi/2,pi,3pi/2,2pi.

    Daniel.
     
  5. Jan 1, 2005 #4

    VietDao29

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    Hi,
    You can use:
    [tex] \tan 4x=\frac{2\tan 2x}{1-\tan^{2}2x} [/tex]
    to solve the problem. Or use this way:
    Since you know that: [tex] \tan{(x + Kpi)} = \tan{x} [/tex]
    And [tex] \tan{4x} = \tan{2x} [/tex]
    <=> [tex] 4x - 2x = Kpi [/tex]
    <=> [tex] 2x = Kpi [/tex]
    <=> [tex] x = K\frac{pi}{2} [/tex]
    K = ...; -3; -2; -1; 0; 1; 2; 3;... (K belongs to Z)
    And the rest you can do.
    Viet Dao,
     
  6. Jan 1, 2005 #5
    ah, i got it, its simply asking for what values of tan2x = 0 in 1 full revolution. Thx.

    so technically you could simply do Tan4x - Tan 2X = tan 2X and then tan2x = 0
     
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