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Math Help - trigonometric equation

  1. #1
    Senior Member pacman's Avatar
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    trigonometric equation

    find the general solution of 2 sin A sin 3A = 1; for A from 0 to 2pi
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  2. #2
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    Quote Originally Posted by pacman View Post
    find the general solution of 2 sin A sin 3A = 1; for A from 0 to 2pi
    2\sin{A} (3\sin{A} - 4\sin^3{A}) = 1

    0 =  8\sin^4{A} - 6\sin^2{A} + 1

    0 = (2\sin^2{A} - 1)(4\sin^2{A} - 1)

    0 = (\sqrt{2}\sin{A} - 1)(\sqrt{2}\sin{A} + 1)(2\sin{A} - 1)(2\sin{A} + 1)
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  3. #3
    Senior Member pacman's Avatar
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    skeeter, the graph shows 8 roots, your answer gave 4 roots, why?
    Attached Thumbnails Attached Thumbnails trigonometric equation-fg.gif  
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  4. #4
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    Quote Originally Posted by pacman View Post
    skeeter, the graph shows 8 roots, your answer gave 4 roots, why?
    each factor has 2 solutions for angle A when set equal to 0 ...

    for example ...

    \sin{A} = \frac{1}{\sqrt{2}}

    A = \frac{\pi}{4} \, , \, \frac{3\pi}{4}
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  5. #5
    Senior Member pacman's Avatar
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    i get it now, its suppliment. Thanks SKEETER
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