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Math Help - trigo problem no. 21

  1. #1
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    trigo problem no. 21

    p195 q27 ex9a

    i've got problem again
    p.s. or is there any other method besides my method 1 and 2 ?
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    p195 q27 ex9a

    i've got problem again
    p.s. or is there any other method besides my method 1 and 2 ?
    I assume 0< x < 360 ......

    In method 1, you've neglected the solutions corresponding to 5x/2 = 360, 540 and 700 ....

    In method 2, if you don't use the calculator then you just have to know the special angles that those exact surd values correspond to ......
    Last edited by mr fantastic; March 13th 2008 at 04:28 AM.
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    Quote Originally Posted by afeasfaerw23231233 View Post
    p195 q27 ex9a

    i've got problem again
    p.s. or is there any other method besides my method 1 and 2 ?
    Method 3 (my favourite):

    sin(2x) = -sin(3x) = sin(-3x).

    Therefore (working in degrees):

    Case 1:

    2x = -3x + 360n, n is an integer

    => 5x = 360n

    => x = 72n

    => x = 0, 72, 144, 216, 288 (assuming 0 < x < 360).

    Case 2:

    2x = 180 - (-3x) + 360n, n is an integer

    => -x = 180 + 360n = 180(1 + 2n)

    => x = -180(2n + 1)

    => x = 180 (assuming 0 < x< 360).
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