Finding solutions to a polynomial by using De Moivre's theorem Use de Moivre’s theorem to show that
$$\ \cos 6θ = 32\cos^6θ − 48\cos^4θ + 18\cos^2θ − 1 $$.
Hence solve the equation
$$\ 64x^6 − 96x^4 + 36x^2 − 1 = 0$$
giving each root in the form$\ \cos kπ$.
Attempt
I have completed almost the whole question and obtained an equation 
$$\ \cos\theta = \cos\frac{k\pi}{9} = -\frac{1}{2}$$
The problem is the equation requires six roots. Whereas the above equation is satisfied by $$\ k= \pm1,\pm2,\pm4,\pm5,\pm7,\pm8$$
My question is 


*

*Which six roots I am supposed to choose? (I am aware that both plus-minus values will give the same value for x, but it is also possible for the same value of x to be the root of the equation twice. Like $\ x=1$ is a root of the equation $\ x^2-2x+1$ twice.)

*Why is such a complication arising?
 A: Assuming you have proved
$$
\cos 6\theta = 32\cos^6\theta − 48\cos^4\theta + 18\cos^2\theta − 1
$$
you get
$$
32\cos^6\theta − 48\cos^4\theta + 18\cos^2\theta=1+\cos6\theta
$$
Now set $x=\cos\theta$, so your equation becomes
$$
2(1+\cos6\theta)-1=0
$$
that is
$$
\cos6\theta=-\frac{1}{2}
$$
which has the solutions
$$
6\theta=\frac{2\pi}{3}+2k\pi
\qquad\text{or}\qquad
6\theta=-\frac{2\pi}{3}+2k\pi
$$
You can choose $0\le k\le 5$, because every other solution will repeat values for $x$. On the other hand, you can discard the second set of solutions, for the same reason, because $\cos(-\alpha)=\cos\alpha$.
So you get
$$
\frac{\pi}{9},\quad
\frac{\pi}{9}+\frac{\pi}{3},\quad
\frac{\pi}{9}+\frac{2\pi}{3},\quad
\frac{\pi}{9}+\frac{3\pi}{3},\quad
\frac{\pi}{9}+\frac{4\pi}{3},\quad
\frac{\pi}{9}+\frac{5\pi}{3}
$$
and the solutions for $x$ are
$$
\cos\frac{\pi}{9},\quad
\cos\frac{4\pi}{9},\quad
\cos\frac{7\pi}{9},\quad
\cos\frac{10\pi}{9},\quad
\cos\frac{13\pi}{9},\quad
\cos\frac{16\pi}{9}
$$
which are pairwise distinct.
To summarize: the equation $\cos6\theta=-1/2$ has twelve solutions in the interval $[0,2\pi)$, but your equation in $x$ has only six, because the solutions in $\theta$ can be grouped in pairs that yield the same value for $\cos\theta$.
