(i) To express \(8 \cos \theta + 15 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identities:

\(\begin{aligned} a \cos \theta + b \sin \theta &\equiv R \cos(\theta - \alpha), \\ R &= \sqrt{a^2 + b^2}, \\ \tan \alpha &= \frac{b}{a}. \end{aligned}\)

Here, \(a = 8\) and \(b = 15\).

Calculate \(R\):

\(R = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{15}{8},\)

\(\alpha = \tan^{-1}\left(\frac{15}{8}\right) \approx 61.93^\circ.\)

(ii) Solve \(8 \cos \theta + 15 \sin \theta = 12\):

Using \(R \cos(\theta - \alpha) = 12\):

\(17 \cos(\theta - 61.93^\circ) = 12,\)

\(\cos(\theta - 61.93^\circ) = \frac{12}{17}.\)

Calculate \(\theta - 61.93^\circ\):

\(\theta - 61.93^\circ = \cos^{-1}\left(\frac{12}{17}\right) \approx 45.099^\circ.\)

Thus, \(\theta = 45.099^\circ + 61.93^\circ = 107.0^\circ\).

Also, \(\theta - 61.93^\circ = 360^\circ - 45.099^\circ = 314.901^\circ\).

Thus, \(\theta = 314.901^\circ + 61.93^\circ = 376.831^\circ\), which is outside the interval \(0^\circ < \theta < 360^\circ\).

Therefore, the solutions are \(\theta = 107.0^\circ\) and \(\theta = 16.8^\circ\).

(a) To express \(5 \sin \theta + 12 \cos \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

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

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

Here, \(a = 5\) and \(b = 12\).

Calculate \(R\):

\(R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{12}{5}\)

\(\alpha = \tan^{-1}\left(\frac{12}{5}\right) \approx 0.395\)

Thus, \(5 \sin \theta + 12 \cos \theta = 13 \cos(\theta - 0.395)\).

(b) Solve \(5 \sin 2x + 12 \cos 2x = 6\) using the result from part (a):

\(5 \sin 2x + 12 \cos 2x = 13 \cos(2x - 0.395)\)

Set \(13 \cos(2x - 0.395) = 6\):

\(\cos(2x - 0.395) = \frac{6}{13}\)

\(2x - 0.395 = \cos^{-1}\left(\frac{6}{13}\right)\)

\(2x = \cos^{-1}\left(\frac{6}{13}\right) + 0.395\)

Calculate \(\cos^{-1}\left(\frac{6}{13}\right)\):

\(\cos^{-1}\left(\frac{6}{13}\right) \approx 1.091\)

\(2x = 1.091 + 0.395\)

\(2x = 1.486\)

\(x = \frac{1.486}{2} \approx 0.743\)

For the second solution:

\(2x = 2\pi - 1.091 + 0.395\)

\(2x = 5.587\)

\(x = \frac{5.587}{2} \approx 2.79\)

Thus, the solutions are \(x = 0.743\) or \(x = 2.79\).

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

\(a \cos x + b \sin x \equiv R \cos(x - \alpha)\)

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

Here, \(a = 1\) and \(b = 3\).

Calculate \(R\):

\(R = \sqrt{1^2 + 3^2} = \sqrt{10}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{3}{1} = 3\)

\(\alpha = \tan^{-1}(3) \approx 71.57^\circ\)

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

Using the result from part (i), express \(\cos 2\theta + 3 \sin 2\theta\) as:

\(\sqrt{10} \cos(2\theta - 71.57^\circ) = 2\)

\(\cos(2\theta - 71.57^\circ) = \frac{2}{\sqrt{10}}\)

\(2\theta - 71.57^\circ = \cos^{-1}\left(\frac{2}{\sqrt{10}}\right)\)

\(2\theta - 71.57^\circ \approx 50.77^\circ\)

\(2\theta \approx 50.77^\circ + 71.57^\circ = 122.34^\circ\)

\(\theta \approx 61.17^\circ\)

For the second solution:

\(2\theta \approx 360^\circ - 50.77^\circ + 71.57^\circ = 380.8^\circ\)

\(\theta \approx 190.4^\circ\)

Since \(\theta\) must be between \(0^\circ\) and \(90^\circ\), the valid solutions are:

\(\theta = 10.4^\circ, 61.2^\circ\)

(i) To express \(\sqrt{6} \cos \theta + \sqrt{10} \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identities:

\(\begin{aligned} a \sin \theta + b \cos \theta &\equiv R \sin(\theta + \alpha), \\ a \cos \theta + b \sin \theta &\equiv R \cos(\theta - \alpha), \\ R &= \sqrt{a^2 + b^2}, \quad \tan \alpha = \frac{b}{a}, \quad 0^\circ < \alpha < 90^\circ. \end{aligned}\)

Here, \(a = \sqrt{6}\) and \(b = \sqrt{10}\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{6})^2 + (\sqrt{10})^2} = \sqrt{6 + 10} = \sqrt{16} = 4.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{\sqrt{10}}{\sqrt{6}} = \sqrt{\frac{10}{6}} = \sqrt{\frac{5}{3}}.\)

\(\alpha = \tan^{-1}(\sqrt{\frac{5}{3}}) \approx 52.24^\circ\).

(ii) (a) Solve \(\sqrt{6} \cos \theta + \sqrt{10} \sin \theta = -4\):

Using \(R \cos(\theta - \alpha) = -4\), we have:

\(\cos(\theta - 52.24^\circ) = \frac{-4}{4} = -1.\)

\(\theta - 52.24^\circ = 180^\circ\), so \(\theta = 180^\circ + 52.24^\circ = 232.24^\circ\).

(b) Solve \(\sqrt{6} \cos \frac{1}{2} \theta + \sqrt{10} \sin \frac{1}{2} \theta = 3\):

Using \(R \cos(\frac{1}{2} \theta - \alpha) = 3\), we have:

\(\cos(\frac{1}{2} \theta - 52.24^\circ) = \frac{3}{4}.\)

\(\frac{1}{2} \theta - 52.24^\circ = \cos^{-1}(0.75) \approx 41.41^\circ\).

\(\frac{1}{2} \theta = 41.41^\circ + 52.24^\circ = 93.65^\circ\).

\(\theta = 2 \times 93.65^\circ = 187.3^\circ\).

Smallest positive \(\theta = 21.7^\circ\).

(i) To express \(5 \sin x + 12 \cos x\) in the form \(R \sin(x + \alpha)\), we use the identity:

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

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

Here, \(a = 5\) and \(b = 12\).

Calculate \(R\):

\(R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{12}{5}\)

\(\alpha = \tan^{-1}\left(\frac{12}{5}\right) \approx 67.38^\circ\)

(ii) Solve \(5 \sin 2\theta + 12 \cos 2\theta = 11\).

Using the result from part (i), express \(5 \sin 2\theta + 12 \cos 2\theta\) as:

\(13 \sin(2\theta + 67.38^\circ) = 11\)

\(\sin(2\theta + 67.38^\circ) = \frac{11}{13}\)

Find \(2\theta + 67.38^\circ\):

\(2\theta + 67.38^\circ = \sin^{-1}\left(\frac{11}{13}\right) \approx 57.80^\circ\)

\(2\theta + 67.38^\circ = 180^\circ - 57.80^\circ = 122.20^\circ\)

Calculate \(2\theta\):

\(2\theta = 57.80^\circ - 67.38^\circ = -9.58^\circ\)

\(2\theta = 122.20^\circ - 67.38^\circ = 54.82^\circ\)

Since \(2\theta\) must be positive, use \(54.82^\circ\):

\(\theta = \frac{54.82^\circ}{2} = 27.41^\circ\)

Find the second solution:

\(2\theta = 180^\circ + 54.82^\circ = 234.82^\circ\)

\(\theta = \frac{234.82^\circ}{2} = 117.41^\circ\)

Thus, the solutions are \(\theta \approx 27.4^\circ, 117.4^\circ\).