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  1. #1
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    tangent

    Solve for x where 0 <= x < 2pi:

    tan(2x) * tan(4x) = 1
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  2. #2
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    Quote Originally Posted by Mr_Green View Post
    Solve for x where 0 <= x < 2pi:

    tan(2x) * tan(4x) = 1
    tan(4x) = tan(2*(2x)) = 2*tan(2x)/[1 - tan^2(2x)]

    So
    tan(2x) * tan(4x) = 1
    becomes

    2*tan^2(2x)/[1 - tan^2(2x)] = 1

    2*tan^2(2x) = 1 - tan^2(2x)

    3*tan^2(2x) = 1

    tan^2(2x) = 1/3

    tan(2x) = 1/sqrt{3} <-- Only keep the + solution since 0 <= x < 2(pi)

    2x = (pi)/6 (rad)

    x = (pi)/12 (rad)

    -Dan
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  3. #3
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    the other values are 5,7,11,17,19,23 pi's over 12 as well
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mr_Green View Post
    the other values are 5,7,11,17,19,23 pi's over 12 as well
    Yes, I didn't give you the full solution set, did I?

    Okay, let's do this the hard way:

    tan(2x)*tan(4x) = 1

    tan(2x) * [2*tan(2x)/(1 - tan^2(2x))] = 1

    2*tan^2(2x)/[1 - tan^2(2x)] = 1

    2*tan^2(2x) = 1 - tan^2(2x)

    3*tan^2(2x) = 1

    So far so good. Now let's expand tan(2x):

    3*[2*tan(x)/(1 - tan^2(x))]^2 = 1

    12*tan^2(x)/(1 - tan^2(x))^2 = 1

    12*tan^2(x) = 1 - 2*tan^2(x) + tan^4(x)

    tan^4(x) - 14*tan^2(x) + 1 = 0

    Let y = tan^2(x). Then:
    y^2 - 14y + 1 = 0

    y = 7 (+/-) 4*sqrt{3}

    So
    tan^2(x) = 7 (+/-) 4*sqrt{3}

    or
    tan(x) = (+/-)sqrt{7 + 4*sqrt{3}} or tan(x) = (+/-)sqrt{7 - 4*sqrt{3}}

    Where to go from here? Well:
    7 + 4*sqrt{3} = (2 + sqrt{3})^2, so
    tan(x) = (+/-)(2 + sqrt{3}) (With a small side of thanks to a recent post of Soroban's )

    and 7 - 4*sqrt{3} = (2 - sqrt{3})^2, so
    tan(x) = (+/-)(2 - sqrt{3})

    This generates the solution set:
    x = (+/-)5(pi)/12, (+/-)(pi)/12

    But we want 0 <= x < 2(pi) so these become:
    x = (pi)/12, 5(pi)/12, 19(pi)/12, and 23(pi)/12

    How do we get the others? Well we may add (pi) to any solution we get from arctangent, so we get the additional solutions:
    x = 13(pi)/12, 17(pi)/12, 31(pi)/12, and 35(pi)/12

    These last two are greater than 2(pi), so we subtract 2(pi) to get
    x = 13(pi)/12, 17(pi)/12, 7(pi)/12, and 11(pi)/12

    Between the two sets of values for x, we have your full solution.

    I guess I shouldn't have taken the short-cut in my first solution.

    -Dan
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  5. #5
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    Hello, Mr_Green!

    Solve for x where 0 < x < 2π

    . . tan(2x)·tan(4x) .= .1

    Using a double-angle identity, we have;

    . . . . . . . . . . 2·tan(2x)
    . . tan(2x) . -------------- · = . 1
    . . . . . . . . .1 - tanČ(2x)


    Then: .2·tanČ(2x) .= .1 - tanČ(2x) . . . . 3·tanČ(2x) .= .1
    . . . . . . . - . . . - . . . . - . . . . . . . . . . . . . . . . . . . . . . _
    And we have: .tanČ(2x) .= .1/3 . . . . tan(2x) .= .±1/√3


    Hence: .2x .= .π/6, 5π/6, 7π/6, 11π/6, 13π/6, 17π/6, 19π/6, 23π/6


    Therefore: .x .= .π/12, 5π/12, 7π/12, 11π/12, 13π/12, 17π/12, 19π/12, 23π/12

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