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Math Help - help solving trig equation is appreciated

  1. #1
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    help solving trig equation is appreciated

    solve: 3sin^2 4x = cos 4x - 1

    given that 0 deg [greater/equal to] x < 360 deg.

    I tried turning it into a quadratic equation using the "1 = sin^2 + cos^2" rule and when that didn't work I pretty much ran out of ideas. Much thanks for any help to be offered.
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  2. #2
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    Quote Originally Posted by Savior_Self View Post
    solve: 3sin^2 4x = cos 4x - 1

    given that 0 deg [greater/equal to] x < 360 deg.

    I tried turning it into a quadratic equation using the "1 = sin^2 + cos^2" rule and when that didn't work I pretty much ran out of ideas. Much thanks for any help to be offered.
    0 \le x < 360

    0 \le 4x < 1440


    3[1 - \cos^2(4x)] = \cos(4x) - 1

    3 - 3\cos^2(4x) = \cos(4x) - 1

    0 = 3\cos^2(4x) + \cos(4x) - 4

    0 = [3\cos(4x) + 4][\cos(4x) - 1]

    \cos(4x) = -\frac{4}{3} ... no solution

    \cos(4x) = 1 ...

    4x = 0

    4x = 360

    4x = 720

    4x = 1080

    you can divide by 4
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  3. #3
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    haha, ahh...I'm an idiot. Thanks skeeter.
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