To solve the equation \(\csc 2\theta = \sec \theta + \cot \theta\), we start by expressing all terms in terms of sine and cosine:

\(\csc 2\theta = \frac{1}{\sin 2\theta} = \frac{1}{2\sin \theta \cos \theta}\)

\(\sec \theta = \frac{1}{\cos \theta}\)

\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)

Substitute these into the equation:

\(\frac{1}{2\sin \theta \cos \theta} = \frac{1}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\)

Multiply through by \(2\sin \theta \cos \theta\) to clear the fractions:

\(1 = 2\sin \theta + 2\cos^2 \theta\)

Using \(\cos^2 \theta = 1 - \sin^2 \theta\), we have:

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

Simplify:

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

\(2\sin^2 \theta - 2\sin \theta - 1 = 0\)

This is a quadratic equation in \(\sin \theta\). Let \(x = \sin \theta\), then:

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

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

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

\(x = \frac{2 \pm \sqrt{4 + 8}}{4}\)

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

\(x = \frac{2 \pm 2\sqrt{3}}{4}\)

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

Thus, \(\sin \theta = \frac{1 + \sqrt{3}}{2}\) or \(\sin \theta = \frac{1 - \sqrt{3}}{2}\).

For \(\sin \theta = \frac{1 + \sqrt{3}}{2}\), solve for \(\theta\):

\(\theta = 201.5^\circ\).

For \(\sin \theta = \frac{1 - \sqrt{3}}{2}\), solve for \(\theta\):

\(\theta = 338.5^\circ\).

Thus, the solutions are \(\theta = 201.5^\circ\) and \(\theta = 338.5^\circ\).