K was found by finding a value such that 27K3/9K = 4/3, so we could turn it into a multiple of the cos3A expansion. This gives K = 2/3.
When we substitute x=2/3cosA in we get 2(cos3A) = 1, so cos3A = 1/2 = cos60, cos300, cos420 etc. (you don't need any more values, since they give the same as those 3 for their respective cosA values).
So A = 20, 100, 140. The roots are 2/3cos20, 2/3cos100, 2/3cos140
From polynomials, you know that the product of the roots = 1/27 (by subtracting 1 from each side, and you get abc = -(-1) where a, b and c are the roots)
so 2/3cos20.2/3cos100.2/3cos140 = 1/27
cos20.cos100.cos140 = 1/8
cos20.cos60.cos100.cos140 = 1/16
When we substitute x=2/3cosA in we get 2(cos3A) = 1, so cos3A = 1/2 = cos60, cos300, cos420 etc. (you don't need any more values, since they give the same as those 3 for their respective cosA values).
So A = 20, 100, 140. The roots are 2/3cos20, 2/3cos100, 2/3cos140
From polynomials, you know that the product of the roots = 1/27 (by subtracting 1 from each side, and you get abc = -(-1) where a, b and c are the roots)
so 2/3cos20.2/3cos100.2/3cos140 = 1/27
cos20.cos100.cos140 = 1/8
cos20.cos60.cos100.cos140 = 1/16