(i) We want to express \(\sqrt{5} \cos x + 2 \sin x\) in the form \(R \cos(x - \alpha)\).

Using the identity:

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

Here, \(a = \sqrt{5}\) and \(b = 2\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{5})^2 + 2^2} = \sqrt{5 + 4} = \sqrt{9} = 3.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{2}{\sqrt{5}},\)

\(\alpha = \tan^{-1}\left(\frac{2}{\sqrt{5}}\right) \approx 41.81^\circ.\)

(ii) Solve \(\sqrt{5} \cos \frac{1}{2}x + 2 \sin \frac{1}{2}x = 1.2\).

We have \(R \cos\left(\frac{1}{2}x - 41.81^\circ\right) = 1.2\).

\(3 \cos\left(\frac{1}{2}x - 41.81^\circ\right) = 1.2\)

\(\cos\left(\frac{1}{2}x - 41.81^\circ\right) = 0.4\)

\(\frac{1}{2}x - 41.81^\circ = \cos^{-1}(0.4) \approx 66.42^\circ\)

\(\frac{1}{2}x = 66.42^\circ + 41.81^\circ = 108.23^\circ\)

\(x = 2 \times 108.23^\circ = 216.46^\circ\)

Rounding to one decimal place, \(x = 216.5^\circ\).

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

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

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

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

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

(i) Expand \(\cos(x + 45^\circ)\):

\(\cos(x + 45^\circ) = \frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}}\)

Express \(\cos(x + 45^\circ) - \sqrt{2} \sin x\):

\(\frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}} - \sqrt{2} \sin x = \frac{\cos x}{\sqrt{2}} - \left(\frac{1}{\sqrt{2}} + \sqrt{2}\right) \sin x\)

Let \(a = \frac{1}{\sqrt{2}}\) and \(b = -\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)\). Then:

\(R = \sqrt{a^2 + b^2} = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)\right)^2} = 2.236\)

\(\tan \alpha = \frac{b}{a} = \frac{-\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)}{\frac{1}{\sqrt{2}}}\)

\(\alpha = \tan^{-1}\left(\frac{-\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)}{\frac{1}{\sqrt{2}}}\right) = 71.57^\circ\)

(ii) Solve \(\cos(x + 45^\circ) - \sqrt{2} \sin x = 2\):

Using the expression from (i):

\(R \cos(x + \alpha) = 2\)

\(2.236 \cos(x + 71.57^\circ) = 2\)

\(\cos(x + 71.57^\circ) = \frac{2}{2.236}\)

\(x + 71.57^\circ = \cos^{-1}\left(\frac{2}{2.236}\right)\)

\(x + 71.57^\circ = 26.57^\circ \text{ or } 333.43^\circ\)

\(x = 315^\circ \text{ or } 261.9^\circ\)

(i) To express \(24 \sin \theta - 7 \cos \theta\) in the form \(R \sin(\theta - \alpha)\), we use the identity:

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

Here, \(a = 24\) and \(b = -7\). Thus,

\(R = \sqrt{24^2 + (-7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25.\)

\(\tan \alpha = \frac{-7}{24}.\)

\(\alpha = \tan^{-1}\left(\frac{-7}{24}\right)\), which gives \(\alpha \approx 16.26^\circ\) (to 2 decimal places).

(ii) To find the smallest positive value of \(\theta\) satisfying \(24 \sin \theta - 7 \cos \theta = 17\), we use:

\(R \sin(\theta - \alpha) = 17.\)

\(25 \sin(\theta - 16.26^\circ) = 17.\)

\(\sin(\theta - 16.26^\circ) = \frac{17}{25}.\)

\(\theta - 16.26^\circ = \sin^{-1}\left(\frac{17}{25}\right)\).

\(\sin^{-1}\left(\frac{17}{25}\right) \approx 42.84^\circ\).

\(\theta = 42.84^\circ + 16.26^\circ = 59.1^\circ\).