Question 980725
If cos^2 x + cos x = 1. so, sin^12 x + 3 sin^10 x + 3 sin^8 x + sin^6 x = ?
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 Solve for cos(x)
 cos^2 x + cos x = 1
 cos^2 x + cos x - 1 = 0
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 Sub u for cos(x)
 u^2 + u - 1 = 0
 *[invoke solve_quadratic_equation 1,1,-1]
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 cos(x) = -1/2 + sqrt(5)/2
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 Find sin^2(x)
 sin^2 + cos^2 = 1
 sin^2(x) = 1 - (1/4 - sqrt(5)/2 + 5/4) = -1/2 - sqrt(5)/2
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 sin^12 x + 3 sin^10 x + 3 sin^8 x + sin^6 x = sin^6(x)*(sin^6 + 3sin^4 + 3sin^2 + 1)
 = sin^6(x)*(sin^2 + 1)^3
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sin^6 = (sin^2)^3 = (1 + sqrt(5))/2
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(1/2)*(1 + sqrt(5))*(-1/2)*(1 - sqrt(5))
= -(1/4)*(1 - 5)
= 1
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Do the same for the other root, cos(x) = -1/2 - sqrt(5)/2