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

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

(a) Start with the given equation:

\(\cos(x - 30^\circ) = 2 \sin(x + 30^\circ)\)

Use the angle subtraction and addition formulas:

\(\cos(x - 30^\circ) = \cos x \cos 30^\circ + \sin x \sin 30^\circ\)

\(\sin(x + 30^\circ) = \sin x \cos 30^\circ + \cos x \sin 30^\circ\)

Substitute the exact values:

\(\cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}\)

\(\cos(x - 30^\circ) = \cos x \frac{\sqrt{3}}{2} + \sin x \frac{1}{2}\)

\(\sin(x + 30^\circ) = \sin x \frac{\sqrt{3}}{2} + \cos x \frac{1}{2}\)

Substitute into the original equation:

\(\cos x \frac{\sqrt{3}}{2} + \sin x \frac{1}{2} = 2 \left( \sin x \frac{\sqrt{3}}{2} + \cos x \frac{1}{2} \right)\)

Expand and simplify:

\(\cos x \frac{\sqrt{3}}{2} + \sin x \frac{1}{2} = \sin x \sqrt{3} + \cos x\)

Rearrange terms:

\(\cos x \frac{\sqrt{3}}{2} - \cos x = \sin x \sqrt{3} - \sin x \frac{1}{2}\)

Factor out terms:

\(\cos x \left( \frac{\sqrt{3}}{2} - 1 \right) = \sin x \left( \sqrt{3} - \frac{1}{2} \right)\)

Divide both sides by \(\cos x\) and \(\sin x\):

\(\tan x = \frac{\frac{\sqrt{3}}{2} - 1}{\sqrt{3} - \frac{1}{2}}\)

Simplify to obtain:

\(\tan x = \frac{2 - \sqrt{3}}{1 - 2\sqrt{3}}\)

(b) From part (a), we have:

\(\tan x = \frac{2 - \sqrt{3}}{1 - 2\sqrt{3}}\)

Calculate \(x\) using the inverse tangent function:

\(x = \tan^{-1}\left(\frac{2 - \sqrt{3}}{1 - 2\sqrt{3}}\right)\)

Find solutions within the interval \(0^\circ < x < 360^\circ\):

\(x \approx 173.8^\circ, 353.8^\circ\)

1. Use the identity \(\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}\) to express \(\tan(x + 45^\circ)\):

\(\tan(x + 45^\circ) = \frac{\tan x + 1}{1 - \tan x}\)

2. Substitute \(\cot x = \frac{1}{\tan x}\) into the equation:

\(\frac{\tan x + 1}{1 - \tan x} = 2 \cdot \frac{1}{\tan x} + 1\)

3. Clear the fractions by multiplying through by \(\tan x (1 - \tan x)\):

\((\tan x + 1) \tan x = 2(1 - \tan x) + \tan x (1 - \tan x)\)

4. Simplify and rearrange to form a quadratic equation:

\(\tan^2 x + \tan x = 2 - 2\tan x + \tan x - \tan^2 x\)

\(2\tan^2 x + \tan x - 1 = 0\)

5. Solve the quadratic equation using the quadratic formula \(\tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(a = 2, \ b = 1, \ c = -1\)

\(\tan x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}\)

\(\tan x = \frac{-1 \pm \sqrt{1 + 8}}{4}\)

\(\tan x = \frac{-1 \pm 3}{4}\)

\(\tan x = \frac{2}{4} = 0.5 \quad \text{or} \quad \tan x = \frac{-4}{4} = -1\)

6. Find \(x\) for \(\tan x = 0.5\):

\(x = \tan^{-1}(0.5) \approx 26.6^\circ\)

7. Find \(x\) for \(\tan x = -1\):

\(x = 135^\circ\)

8. Adjust for the given interval \(0^\circ < x < 180^\circ\):

\(x = 31.7^\circ \quad \text{and} \quad x = 121.7^\circ\)

(a) Start with the identity for tangent of a sum:

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \tan 60^\circ}{1 - \tan \theta \tan 60^\circ}\)

Since \(\tan 60^\circ = \sqrt{3}\), substitute to get:

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}}\)

Given \(\tan(\theta + 60^\circ) = 2 \cot \theta\), rewrite \(\cot \theta\) as \(\frac{1}{\tan \theta}\):

\(\frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}} = \frac{2}{\tan \theta}\)

Cross-multiply to obtain:

\(\tan \theta (\tan \theta + \sqrt{3}) = 2(1 - \tan \theta \sqrt{3})\)

Expand and simplify:

\(\tan^2 \theta + \sqrt{3} \tan \theta = 2 - 2\sqrt{3} \tan \theta\)

Rearrange to form the quadratic equation:

\(\tan^2 \theta + 3\sqrt{3} \tan \theta - 2 = 0\)

(b) Solve the quadratic equation \(\tan^2 \theta + 3\sqrt{3} \tan \theta - 2 = 0\) using the quadratic formula:

\(\tan \theta = \frac{-3\sqrt{3} \pm \sqrt{(3\sqrt{3})^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}\)

Calculate the discriminant:

\(\sqrt{27 + 8} = \sqrt{35}\)

Thus,

\(\tan \theta = \frac{-3\sqrt{3} \pm \sqrt{35}}{2}\)

Calculate the solutions for \(\theta\):

\(\tan \theta = 0.3599\) gives \(\theta = 19.8^\circ\)

\(\tan \theta = -5.556\) gives \(\theta = 100.2^\circ\)

Thus, the solutions are \(\theta = 19.8^\circ\) and \(\theta = 100.2^\circ\).