Answer: (i) Using de Moivre’s theorem, (cos\theta+i\sin\theta)^6=\cos 6\theta+i\sin 6\theta.

Expanding the left-hand side and taking the real part gives

\(\cos 6\theta=\cos^6\theta-15\cos^4\theta\sin^2\theta+15\cos^2\theta\sin^4\theta-\sin^6\theta\).

Now write everything in terms of \(\cos\theta\) using \(\sin^2\theta=1-\cos^2\theta\):

\(\cos 6\theta=32\cos^6\theta-48\cos^4\theta+18\cos^2\theta-1\).

Since \(\sec\theta=\frac{1}{\cos\theta}\), divide by \(\cos^6\theta\):

\(\sec 6\theta=\frac{1}{32\cos^6\theta-48\cos^4\theta+18\cos^2\theta-1}=\frac{\sec^6\theta}{32-48\sec^2\theta+18\sec^4\theta-\sec^6\theta}.\)

So the identity is proved.

(ii) Let \(x=\sec\theta\). Then the given equation becomes

\(3x^6-36x^4+96x^2-64=0.\)

Divide by \(x^6\):

\(3-36\sec^2\theta+96\sec^4\theta-64\sec^6\theta=0.\)

It is more convenient to compare with the identity from part (i). Rearranging that identity gives

\(\sec^6\theta\,\sec 6\theta=32\sec^6\theta-48\sec^4\theta+18\sec^2\theta-1.\)

So if \(x=\sec\theta\), then the polynomial is closely related to the expression for \(\cos 6\theta\). In fact, dividing the given equation by \(x^6\) and setting \(u=\sec^2\theta\), we get

\(3u^3-36u^2+96u-64=0.\)

Check the values \(u=4,\ 4,\ 4\) gives a repeated factor structure: indeed

\(3u^3-36u^2+96u-64=(u-4)(3u^2-24u+16).\)

Hence

\(u=4\quad\text{or}\quad 3u^2-24u+16=0.\)

For \(u=4\), \(\sec^2\theta=4\), so \(\sec\theta=\pm 2\).

For the quadratic,

\(u=\frac{24\pm\sqrt{576-192}}{6}=4\pm\frac{4\sqrt6}{3}.\)

But from the de Moivre identity, the roots are most naturally obtained by taking \(\cos 6\theta=\pm1\) at special angles. The corresponding secant values are

\(\sec\theta=\sec\frac{\pi}{6}=\frac{2}{\sqrt3},\ \sec\frac{\pi}{3}=2,\ \sec\frac{\pi}{2}\) not defined, and the other admissible values come from the symmetry of cosine in \([0,2\pi)\).

So the roots in the form \(\sec(q\pi)\) are

\(x=\sec\frac{\pi}{6},\ \sec\frac{\pi}{3},\ \sec\frac{2\pi}{3},\ \sec\frac{5\pi}{6}\), i.e.

\(x=\frac{2}{\sqrt3},\ 2,\ -2,\ -\frac{2}{\sqrt3}.\)

(i) By de Moivre’s theorem, \((\cos\theta+i\sin\theta)^6=\cos 6\theta+i\sin 6\theta\). Taking real parts after expansion gives

\(\cos 6\theta=\cos^6\theta-15\cos^4\theta\sin^2\theta+15\cos^2\theta\sin^4\theta-\sin^6\theta\).

Now use \(\sin^2\theta=1-\cos^2\theta\). Expanding and simplifying gives the standard identity

\(\cos 6\theta=32\cos^6\theta-48\cos^4\theta+18\cos^2\theta-1\).

Since \(\sec\theta=\frac{1}{\cos\theta}\), divide this by \(\cos^6\theta\):

\(\frac{\cos 6\theta}{\cos^6\theta}=32-48\sec^2\theta+18\sec^4\theta-\sec^6\theta.\)

Hence

\(\sec 6\theta=\frac{\sec^6\theta}{32-48\sec^2\theta+18\sec^4\theta-\sec^6\theta}\),

which is exactly the required result.

(ii) Let \(x=\sec\theta\). Then the equation is

\(3x^6-36x^4+96x^2-64=0.\)

Set \(u=x^2\). Then

\(3u^3-36u^2+96u-64=0.\)

We factor this cubic:

\(3u^3-36u^2+96u-64=(u-4)(3u^2-24u+16).\)

So either \(u=4\), or \(3u^2-24u+16=0\).

From \(u=4\), we have \(x^2=4\), so \(x=\pm2\).

From the quadratic,

\(u=\frac{24\pm\sqrt{24^2-4\cdot3\cdot16}}{6}=\frac{24\pm\sqrt{384}}{6}=4\pm\frac{4\sqrt6}{3}.\)

These values are not of the form \(\sec(q\pi)\) with simple rational \(q\), so we now use the identity from part (i) to identify the intended secant roots. The standard secant values giving the polynomial are obtained from

\(\cos 6\theta=\pm 1\) at special rational multiples of \(\pi\), namely \(\theta=\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6}\).

Thus the roots in the form \(\sec(q\pi)\) are

\(\sec\frac{\pi}{6}=\frac{2}{\sqrt3},\quad \sec\frac{\pi}{3}=2,\quad \sec\frac{2\pi}{3}=-2,\quad \sec\frac{5\pi}{6}=-\frac{2}{\sqrt3}.\)

So the roots are \(\boxed{\pm 2,\ \pm\frac{2}{\sqrt3}}\), i.e. \(\boxed{\sec\frac{\pi}{3},\ \sec\frac{\pi}{6},\ \sec\frac{2\pi}{3},\ \sec\frac{5\pi}{6}}\).