(i) To express \(3 \sin \theta + 2 \cos \theta\) in the form \(R \sin(\theta + \alpha)\), we use the identity:

\(a \sin \theta + b \cos \theta \equiv R \sin(\theta + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 3\) and \(b = 2\).

Calculate \(R\):

\(R = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{2}{3}\)

\(\alpha = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.69^\circ\)

(ii) Solve \(3 \sin \theta + 2 \cos \theta = 1\):

Using the expression from part (i):

\(R \sin(\theta + \alpha) = 1\)

\(\sqrt{13} \sin(\theta + 33.69^\circ) = 1\)

\(\sin(\theta + 33.69^\circ) = \frac{1}{\sqrt{13}}\)

\(\theta + 33.69^\circ = \sin^{-1}\left(\frac{1}{\sqrt{13}}\right)\)

\(\sin^{-1}\left(\frac{1}{\sqrt{13}}\right) \approx 16.10^\circ\)

\(\theta + 33.69^\circ = 16.10^\circ\) or \(\theta + 33.69^\circ = 180^\circ - 16.10^\circ\)

\(\theta = 16.10^\circ - 33.69^\circ = -17.59^\circ\) (not in range)

\(\theta = 180^\circ - 16.10^\circ - 33.69^\circ = 130.21^\circ\)

Thus, \(\theta = 130.2^\circ\).