Answer: (i) \(\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta\)

(ii) The roots are \(x = \tan\frac{\pi}{8},\; \tan\frac{3\pi}{8},\; \tan\frac{5\pi}{8},\; \tan\frac{7\pi}{8}\)

(i) By de Moivre's theorem,

\((\cos\theta + i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta\).

Now expand the left-hand side:

\((\cos\theta + i\sin\theta)^4 = \cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + i^4\sin^4\theta\).

Since \(i^4=1\), the real part is

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

Equating real parts gives

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

as required.

(ii) Let \(x = \tan\theta\). Then

\(x^4 - 6x^2 + 1 = 0\)

becomes

\(\tan^4\theta - 6\tan^2\theta + 1 = 0\).

Divide the identity in part (i) by \(\cos^4\theta\):

\(\cos 4\theta = \frac{\cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta}{\cos^4\theta}\cos^4\theta\).

More usefully, write everything in terms of \(\tan\theta\):

\(\cos 4\theta = \frac{1 - 6\tan^2\theta + \tan^4\theta}{(1+\tan^2\theta)^2}\).

So if \(x = \tan\theta\), then the equation \(x^4 - 6x^2 + 1 = 0\) is equivalent to

\(\cos 4\theta = 0\).

Hence

\(4\theta = \frac{\pi}{2} + k\pi\), where \(k\in\mathbb{Z}\).

Therefore

\(\theta = \frac{\pi}{8} + k\frac{\pi}{4}\).

Taking all distinct values of \(\tan\theta\), we get

\(x = \tan\frac{\pi}{8},\; \tan\frac{3\pi}{8},\; \tan\frac{5\pi}{8},\; \tan\frac{7\pi}{8}\).

These are the roots in the form \(\tan q\pi\) with positive rational \(q\). Equivalently, since \(\tan(\pi-\alpha)=-\tan\alpha\), the four roots are \(\pm\tan\frac{\pi}{8}\) and \(\pm\tan\frac{3\pi}{8}\).