(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, \(\sin^2 \theta = \frac{1}{4}\).

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

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