(i) Start with the given equation:

\(\sec \theta + 2 \csc \theta = 3 \csc 2\theta\)

Convert to sines and cosines:

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

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

\(2 \sin \theta + 4 \cos \theta = 3\)

(ii) Express \(2 \sin \theta + 4 \cos \theta\) in the form \(R \sin(\theta + \alpha)\):

\(R = \sqrt{2^2 + 4^2} = \sqrt{20} \approx 4.47\)

\(\tan \alpha = \frac{4}{2} = 2\)

\(\alpha = \tan^{-1}(2) \approx 63.43^\circ\)

(iii) Solve \(\sec \theta + 2 \csc \theta = 3 \csc 2\theta\):

Using the result from (i), solve \(2 \sin \theta + 4 \cos \theta = 3\):

\(\sin(\theta + 63.43^\circ) = \frac{3}{4.47}\)

\(\theta + 63.43^\circ = \sin^{-1}\left(\frac{3}{4.47}\right)\)

\(\theta = 74.4^\circ, 338.7^\circ\)