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Math Help - Solve for all angles [0,2pi]

  1. #1
    Newbie
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    Jun 2005
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    Solve for all angles [0,2pi] HELP!! PLEASE

    how do i solve for all angles [0,2pi]????

    sin2x + sin x = 0



    cos^2-cos 2x = 0



    PLEASE HELP!!! THANKS
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  2. #2
    MHF Contributor
    Joined
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    You expand the equations in terms of angle x only. Meaning, expand sin(2x) and cos(2x) into trig functions/function of x only. The idea to see if you can form equations in x only that you can solve algebraocally.
    [Say what? :-) . Okay, it is really easier to explain with numbers...]

    sin(2x) +sin(x) = 0
    2sin(x)cos(x) +sin(x) = 0
    [sin(x)]*[2cos(x) +1] = 0

    sin(x) = 0
    x = arcsin(0)
    x = 0, pi, and 2pi ---***

    2cos(x) +1 = 0
    2cos(x) = -1
    cos(x) = -1/2
    x = arccos(-1/2)
    x is in the 2nd and 3rd quadrant, arccos(1/2) is 60 degrees or pi/3, so,
    In the 2nd quadrant, x = pi -pi/3 = 2pi/3 ---***
    In the 3rd quadrant, x = pi +pi/3 = 4pi/3 ----***

    Therefore, in the interval [0,2pi],
    x = 0, 2pi/3, pi, 4pi/3, 2pi --------------answer.

    ------------------------
    cos^2(x) -cos(2x) = 0
    cos^2(x) -[cos^2(x) -sin^2(x)] = 0
    sin^2(x) = 0
    sin(x) = 0
    x = arcsin(0)
    x = 0, pi, 2pi ----answer.

    -----------------
    To see if those answers are correct, try plugging them into the original equations. If they make the original equations true, then the answers are correct.

    [Hint: They are correct.]
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  3. #3
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    Jun 2005
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    Thanks

    Thank You Very Much!!!
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