Solution: Given that \(e^{i \theta} = \cos \theta + i \sin \theta\) we must have that
- \begin{align*}
\cos 4 \theta &= \textrm{Re} \l e^{i 4 \theta} \r \\
&= \textrm{Re} \l (\cos \theta + i \sin \theta)^4 \r \\
&= \cos^4 \theta - \binom{4}{2}\cos^2 \theta \sin^2 \theta +\sin^4 \theta \\
&= \cos^4 \theta - 6\cos^2 \theta
(1-\cos^2 \theta) +(1-\cos^2 \theta)^2 \\
&= 8\cos^4 \theta - 8\cos^2 \theta + 1
\end{align*}
- Similarly,
\begin{align*}
\cos 6 \theta &= \textrm{Re} \l e^{i 6 \theta} \r \\
&= \textrm{Re} \l (\cos \theta + i \sin \theta)^6 \r \\
&= \cos^6 \theta -\binom{6}{2}\cos^4 \theta \sin^2 \theta +\binom{6}{4} \cos^2\theta \sin^4 \theta - \sin^6 \theta \\
&= \cos^6 \theta - 15 \cos^4 \theta (1-\cos^2 \theta) + 15\cos^2 \theta (1-\cos^2\theta)^2 - (1-\cos^2 \theta)^3\\
&= 31\cos^6 \theta-45\cos^4\theta+15\cos^2\theta-1+3\cos^2 \theta-3\cos^4 \theta+\cos^6 \theta \\
&= 32 \cos^6 \theta-48\cos^4 \theta+18\cos^2 \theta-1
\end{align*}
\begin{align*}
0 &= 16x^{6}-28x^{4}+13x^{2}-1\\
&= \frac12 (32x^6-56x^4+26x^2-1) \\
&= \frac12(32x^6-48x^4+18x^2-1-(8x^4-8x^2+1))
\end{align*}
Therefore if \(x = \cos \theta\) then we are looking at solving \(\cos 6 \theta = \cos 4 \theta\).
\(\cos 6 \theta - \cos 4 \theta = -2 \sin 5\theta \sin \theta = 0\). So we should be looking at \(\sin 5 \theta = 0\) and \(\sin \theta = 0\).
\(\sin \theta = 0 \Rightarrow x = \cos \theta = \pm 1\) both of which are roots.
The other roots will be \(\cos \frac{\pi}{5}, \cos \frac{2\pi}{5}\) etc but it's unclear this is an acceptable form.
Alternatively, given our two roots, we can factorize
\begin{align*}
0 &= 16x^{6}-28x^{4}+13x^{2}-1 \\
&= (x^2-1)(16x^4-12x^2+1)
\end{align*}
We can solve \(16y^2-12y+1=0\) to see that \(x^2 = \frac{3 \pm \sqrt{5}}{8}\) so our roots are:
\(x = -1, 1, \pm \sqrt{\frac{3 + \sqrt{5}}{8}}, \pm \sqrt{\frac{3 -\sqrt{5}}{8}}\)
(We might notice that \(3+\sqrt{5} =\l \frac{1+\sqrt{5}}{\sqrt{2}} \r^2\) so our final answer could be: \(x = -1, 1, \pm \frac{1+\sqrt{5}}{4}, \pm \frac{\sqrt{5}-1}{4}\))