(a) Start by using the double angle formula:

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

Using \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\), substitute:

\(\tan 4\theta = \frac{2 \left( \frac{2 \tan \theta}{1 - \tan^2 \theta} \right)}{1 - \left( \frac{2 \tan \theta}{1 - \tan^2 \theta} \right)^2}\)

Simplify to get:

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

Set \(\tan 4\theta = \frac{1}{2} \tan \theta\) and simplify:

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

Cancel \(\tan \theta\) and solve:

\(8 (1 - \tan^2 \theta) = (1 - \tan^2 \theta)^2 - 4 \tan^2 \theta\)

Rearrange and simplify to get:

\(\tan^4 \theta + 2 \tan^2 \theta - 7 = 0\)

(b) Solve \(\tan^4 \theta + 2 \tan^2 \theta - 7 = 0\) by substituting \(x = \tan^2 \theta\):

\(x^2 + 2x - 7 = 0\)

Use the quadratic formula:

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-7)}}{2 \times 1}\)

\(x = \frac{-2 \pm \sqrt{4 + 28}}{2}\)

\(x = \frac{-2 \pm \sqrt{32}}{2}\)

\(x = \frac{-2 \pm 4\sqrt{2}}{2}\)

\(x = -1 \pm 2\sqrt{2}\)

Only positive \(x\) is valid: \(x = -1 + 2\sqrt{2}\)

\(\tan^2 \theta = -1 + 2\sqrt{2}\)

\(\tan \theta = \sqrt{-1 + 2\sqrt{2}}\)

Calculate \(\theta\) for \(0^\circ < \theta < 180^\circ\):

\(\theta = 53.5^\circ, 126.5^\circ\)