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!!

You should have the four pairs of complex roots you have accounted for plus a factor $x-1$ as $x=1$ is a root of your polynomial.  That gives you the linear term.
