(a) Expand \((2x - 1)(4x^2 + 2x - 1)\):

\((2x - 1)(4x^2 + 2x - 1) = 8x^3 + 4x^2 - 2x - 4x^2 - 2x + 1 = 8x^3 - 4x + 1\).

Thus, the identity is verified.

(b) Start with the left-hand side:

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

Thus, the identity is proven.

(c) Using the results of (a) and (b), solve:

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

Cross-multiply to get:

\(1 = 4\cos \theta (1 - 2\cos^2 \theta)\).

Simplify to:

\(4\cos \theta - 8\cos^3 \theta = 1\).

Let \(x = \cos \theta\), then:

\(8x^3 - 4x + 1 = 0\).

Using the identity from (a), solve for \(x\):

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

\(2x - 1 = 0\) gives \(x = \frac{1}{2}\), \(\cos \theta = \frac{1}{2}\), \(\theta = 60^\circ\).

\(4x^2 + 2x - 1 = 0\) gives \(x = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm 4}{8}\).

\(x = \frac{1}{4}\), \(\cos \theta = \frac{1}{4}\), \(\theta = 75.5^\circ\) (not in range).

\(x = -\frac{3}{4}\), \(\cos \theta = -\frac{3}{4}\), \(\theta = 144^\circ\).

Thus, \(\theta = 60^\circ, 72^\circ, 144^\circ\).