<s>[INST] using $x^9 - 1$ as a product of linear and quadratic factors with real coefficients to solve trig So here's where I'm at I know i can find the ninth roots of unity by using the nth roots of unity formula  $$ \zeta_n = e^{\frac{i2\pi k}{n}} ,$$ when $k = 1,2,3,...,n-1$ 
so the roots of unity where $e^{i\frac{2\pi}{9}},e^{i\frac{4\pi}{9}},...,e^{i\frac{8\pi}{9}}$and the other 4 would be the same except the angles would just be negative.
The previous parts of the question was to show that $$(x-e^{i\theta})(x-e^{-i\theta})$$ could be expressed as a quadratic with real coefficients the answer I got was $$(x-e^{i\theta})(x-e^{-i\theta})=x^2-2x\cos(\theta)+1$$so I said well the $x^9 - 1= (x-e^{i\frac{2\pi}{9}})(x-e^{-i\frac{2\pi}{9}})...(x-e^{i\frac{8\pi}{9}})(x-e^{-i\frac{8\pi}{9}})(x-1)$ and then using the formula I proved said that, that was equal to $$(x^2-2x\cos(\frac{2\pi}{9})+1)(x^2-2x\cos(\frac{4\pi}{9})+1)...(x^2+2x\cos(\frac{8\pi}{9})+1)(x-1)$$
Somehow I need to use this to solve 
$$\cos(\frac{2\pi}{9})+\cos(\frac{4\pi}{9})+\cos(\frac{6\pi}{9})+\cos(\frac{8\pi}{9})= -\frac{1}{2}$$
Any help ,much appreciated!!
 [/INST] Note sure where this problem came from or what kind of solution is expected, but this runs up against the Casus Irreducibilus. Let $\zeta=\zeta_9$ be an irreducible $9$th root of unity, then you want the polynomial
$$(x-\zeta)(x-\zeta^{-1})=x^2-(\zeta+\zeta^{-1})x+1$$
So to find the coefficient we have 
$$(\zeta+\zeta^{-1})^3=3(\zeta+\zeta^{-1})+(\zeta^3+\zeta^{-3})$$
So $\zeta+\zeta^{-1}$ is the solution of 
$$x^3-3x+1=0.$$
Now the discriminant is $D=3^4$ meaning that the roots are real, but cannot be expressed as cube roots of real numbers. Using Cardano's formula
$$\zeta+\zeta^{-1}=\sqrt[3]{\frac{-1+\sqrt{-3}}{2}}+\sqrt[3]{\frac{-1-\sqrt{-3}}{2}}$$
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