(a) Start with the given equation:

\(\cot^2 \theta + 2 \cos 2\theta = 4\)

Use the identity \(\cot^2 \theta = \csc^2 \theta - 1\):

\(\csc^2 \theta - 1 + 2 \cos 2\theta = 4\)

Substitute \(\csc^2 \theta = \frac{1}{\sin^2 \theta}\) and \(\cos 2\theta = 1 - 2 \sin^2 \theta\):

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

Simplify:

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

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

Multiply through by \(\sin^2 \theta\):

\(1 - 4 \sin^4 \theta + \sin^2 \theta = 4 \sin^2 \theta\)

Rearrange to:

\(4 \sin^4 \theta + 3 \sin^2 \theta - 1 = 0\)

(b) Solve the quadratic equation:

Let \(x = \sin^2 \theta\), then:

\(4x^2 + 3x - 1 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(x = \frac{-3 \pm \sqrt{3^2 - 4 \times 4 \times (-1)}}{2 \times 4}\)

\(x = \frac{-3 \pm \sqrt{9 + 16}}{8}\)

\(x = \frac{-3 \pm 5}{8}\)

\(x = \frac{2}{8} = \frac{1}{4} \quad \text{or} \quad x = \frac{-8}{8} = -1\)

Since \(x = \sin^2 \theta\), \(x = -1\) is not valid. So, \(\sin^2 \theta = \frac{1}{4}\).

\(\sin \theta = \pm \frac{1}{2}\)

\(\theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ\)

To solve \(\cot 2\theta = 2 \tan \theta\), we start by using trigonometric identities:

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

Equating this to \(2 \tan \theta\):

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

Cross-multiply to clear the fraction:

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

Rearrange the equation:

\(1 = 5 \tan^2 \theta\)

\(\tan^2 \theta = \frac{1}{5}\)

Take the square root:

\(\tan \theta = \pm \frac{1}{\sqrt{5}}\)

Calculate \(\theta\) using \(\tan^{-1}\):

\(\theta = \tan^{-1} \left( \frac{1}{\sqrt{5}} \right) \approx 24.1^\circ\)

For the second solution in the range \(0^\circ < \theta < 180^\circ\):

\(\theta = 180^\circ - 24.1^\circ = 155.9^\circ\)

Thus, the solutions are \(\theta = 24.1^\circ\) and \(\theta = 155.9^\circ\).

(i) Start with the identity \(\cos^2 x + \sin^2 x = 1\). Raise both sides to the power of 3:

\((\cos^2 x + \sin^2 x)^3 = 1^3 = 1\).

Expand \((\cos^2 x + \sin^2 x)^3\):

\((\cos^2 x + \sin^2 x)^3 = \cos^6 x + 3\cos^4 x \sin^2 x + 3\cos^2 x \sin^4 x + \sin^6 x\).

Using \(\cos^2 x + \sin^2 x = 1\), substitute \(\cos^2 x = 1 - \sin^2 x\) and \(\sin^2 x = 1 - \cos^2 x\):

\(\cos^6 x + \sin^6 x = 1 - 3\cos^2 x \sin^2 x\).

Using the double angle identity \(\sin 2x = 2 \sin x \cos x\), we have \(\sin^2 2x = 4 \sin^2 x \cos^2 x\):

\(\cos^6 x + \sin^6 x = 1 - \frac{3}{4} \sin^2 2x\).

(ii) Given \(\cos^6 x + \sin^6 x = \frac{2}{3}\), substitute the result from part (i):

\(1 - \frac{3}{4} \sin^2 2x = \frac{2}{3}\).

Solve for \(\sin^2 2x\):

\(\frac{3}{4} \sin^2 2x = 1 - \frac{2}{3} = \frac{1}{3}\).

\(\sin^2 2x = \frac{4}{9}\).

\(\sin 2x = \pm \frac{2}{3}\).

For \(\sin 2x = \frac{2}{3}\), solve for \(2x\):

\(2x = \sin^{-1}\left(\frac{2}{3}\right)\) gives \(2x \approx 41.8^\circ\) or \(2x \approx 138.2^\circ\).

Thus, \(x \approx 20.9^\circ\) or \(x \approx 69.1^\circ\).

For \(\sin 2x = -\frac{2}{3}\), solve for \(2x\):

\(2x = \sin^{-1}\left(-\frac{2}{3}\right)\) gives \(2x \approx 221.8^\circ\) or \(2x \approx 318.2^\circ\).

Thus, \(x \approx 110.9^\circ\) or \(x \approx 159.1^\circ\).

Therefore, the solutions are \(x = 20.9^\circ, 69.1^\circ, 110.9^\circ, 159.1^\circ\).

Step 1: Express \(\cot \theta\) and \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).

\(\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}\)

Step 2: Substitute into the equation:

\(\frac{\cos \theta}{\sin \theta} - 2 \left( \frac{\sin \theta}{\cos \theta} \right) = \sin 2\theta\)

Step 3: Use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\):

\(\frac{\cos^2 \theta - 2 \sin^2 \theta}{\sin \theta \cos \theta} = 2 \sin \theta \cos \theta\)

Step 4: Multiply through by \(\sin \theta \cos \theta\) to clear the fraction:

\(\cos^2 \theta - 2 \sin^2 \theta = 2 \sin^2 \theta \cos^2 \theta\)

Step 5: Use \(\sin^2 \theta = 1 - \cos^2 \theta\):

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

Step 6: Simplify:

\(\cos^2 \theta - 2 + 2 \cos^2 \theta = 2 \cos^2 \theta - 2 \cos^4 \theta\)

Step 7: Rearrange to form a quadratic in \(\cos^2 \theta\):

\(2 \cos^4 \theta + \cos^2 \theta - 2 = 0\)

Step 8: Solve for \(\cos^2 \theta\):

Let \(x = \cos^2 \theta\), then:

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

Step 9: Solve the quadratic equation using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

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

\(x = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4}\)

Step 10: Find \(\theta\) for \(90^\circ < \theta < 180^\circ\):

Only the positive root is valid for \(\cos^2 \theta\), so:

\(\cos^2 \theta = \frac{-1 + \sqrt{17}}{4}\)

\(\theta = 152.1^\circ\)

To express \(\cot 2\theta\) in terms of \(\tan \theta\), use the identity:

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

Given the equation:

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

Multiply through by \(2\tan \theta\) to clear the fraction:

\(1 - \tan^2 \theta = 2\tan \theta + 2\tan^2 \theta\)

Rearrange to form a quadratic equation:

\(3\tan^2 \theta + 2\tan \theta - 1 = 0\)

Let \(x = \tan \theta\). The equation becomes:

\(3x^2 + 2x - 1 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = 2\), \(c = -1\):

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

\(x = \frac{-2 \pm \sqrt{4 + 12}}{6}\)

\(x = \frac{-2 \pm \sqrt{16}}{6}\)

\(x = \frac{-2 \pm 4}{6}\)

\(x = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = \frac{-6}{6} = -1\)

Thus, \(\tan \theta = \frac{1}{3}\) or \(\tan \theta = -1\).

For \(\tan \theta = \frac{1}{3}\), \(\theta = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.4^\circ\).

For \(\tan \theta = -1\), \(\theta = 135^\circ\) (since \(\tan 135^\circ = -1\)).

Thus, the solutions are \(\theta = 18.4^\circ\) and \(\theta = 135^\circ\).