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u^3+3u=39-3cos(t)

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  • 5 years ago
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    Let u = v cos(r)

    v^3 cos^3(r) + 3 v cos(r) = 39 - 3cos(t)

    Multiply by 4 and divide by v^3

    4 cos^3(r) + (12/v^2) cos(r) + [ 3cos(t) - 39 ] (12/v^3) = 0

    Define v such that (12/v^2) = -3

    v^2 = -4

    v = 2i

    and

    4 cos^3(r) - 3 cos(r) + [ 3cos(t) - 39 ] (3i/2) = 0

    Now let cos(3r) = [ 39 - 3cos(t) ] (3i/2)

    3r = 2pi n + arccos([ 39 - 3cos(t) ] (3i/2))

    r = (2/3)pi n + (1/3)arccos([ 39 - 3cos(t) ] (3i/2))

    Now you know that u = v cos(r)

    So

    u = 2i cos[ (2/3)pi n + (1/3)arccos([ 39 - 3cos(t) ] (3i/2)) ]

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