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# solve for u?

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

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- 5 years agoFavorite Answer
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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