HMMMM ok...
So continuing from where you left off at:
\(\cos^4 \beta + \cos^2 \beta = 1 \\ (\cos^4 \beta + \cos^2 \beta)^2 = 1^2 \\ \cos^8 \beta + 2\cos^6 \beta + \cos^4 \beta = 1 \\ \cos^8 \beta + 2\cos^6 \beta + \cos^4 \beta - 1 = 0\)
And that matches this form: \(a\cos^8 \beta + b\cos^6 \beta + c \cos^4 \beta -1 = 0\)
Which means a = 1, b = 2, and c = 1
Next:
\(\cos^4 \beta + \cos^2 \beta = 1 \\ (\cos^4 \beta + \cos^2 \beta)^3 = 1^3 \\ \cos^{12} \beta + 3cos^{10} \beta +3\cos^8 \beta + \cos^6 \beta= 1 \\ \cos^{12} \beta + 3cos^{10} \beta +3\cos^8 \beta + \cos^6 \beta - 1 =0\)
That looks almost exactly like: \(d\cos^{12} \beta + e\cos^{10} \beta + f\cos^8 \beta + g\cos^6 \beta = 0\)
But its not exactly the same because there's a - 1 in one and not in the other.
So now I'm stuck again.