(i) To express \(7 \cos \theta + 24 \sin \theta\) 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, \(a = 7\) and \(b = 24\).

\(R = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25.\) \(\tan \alpha = \frac{24}{7} \implies \alpha = \tan^{-1}\left(\frac{24}{7}\right) \approx 73.74^\circ.\)

(ii) To solve \(7 \cos \theta + 24 \sin \theta = 15\), we use the expression:

\(R \cos(\theta - \alpha) = 15.\)

Substitute \(R = 25\) and \(\alpha = 73.74^\circ\):

\(25 \cos(\theta - 73.74^\circ) = 15 \implies \cos(\theta - 73.74^\circ) = \frac{15}{25} = 0.6.\)

Find \(\theta - 73.74^\circ\):

\(\theta - 73.74^\circ = \cos^{-1}(0.6) \approx 53.13^\circ.\)

Thus, \(\theta = 53.13^\circ + 73.74^\circ = 126.87^\circ\).

Also, \(\theta - 73.74^\circ = 360^\circ - 53.13^\circ = 306.87^\circ\).

Thus, \(\theta = 306.87^\circ + 73.74^\circ = 20.61^\circ\).

Therefore, the solutions are \(\theta = 126.9^\circ\) and \(\theta = 20.6^\circ\).

First, express \(8 \sin \theta - 6 \cos \theta\) in the form \(R \sin(\theta - \alpha)\).

Using 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, \(a = 8\) and \(b = -6\).

Calculate \(R\):

\(R = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-6}{8} = -\frac{3}{4}.\)

\(\alpha = \tan^{-1}\left(-\frac{3}{4}\right) \approx -36.87^\circ.\)

Thus, \(8 \sin \theta - 6 \cos \theta = 10 \sin(\theta + 36.87^\circ)\).

Now, solve the equation:

\(10 \sin(\theta + 36.87^\circ) = 7.\)

\(\sin(\theta + 36.87^\circ) = 0.7.\)

Find \(\theta + 36.87^\circ\):

\(\theta + 36.87^\circ = \sin^{-1}(0.7) \approx 44.427^\circ.\)

\(\theta = 44.427^\circ - 36.87^\circ = 7.557^\circ.\)

Considering the range \(0^\circ \leq \theta \leq 360^\circ\), find the other solution:

\(\theta = 180^\circ - 44.427^\circ - 36.87^\circ = 172.4^\circ.\)

Thus, the solutions are \(\theta = 81.3^\circ\) and \(\theta = 172.4^\circ\).

(i) To express \(4 \sin \theta - 3 \cos \theta\) 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, \(a = 4\) and \(b = 3\).

Calculate \(R\):

\(R = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{3}{4}\)

\(\alpha = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.87^\circ\)

(ii) Solve \(4 \sin \theta - 3 \cos \theta = 2\):

Using the form \(R \sin(\theta - \alpha)\), we have:

\(5 \sin(\theta - 36.87^\circ) = 2\)

\(\sin(\theta - 36.87^\circ) = \frac{2}{5}\)

\(\theta - 36.87^\circ = \sin^{-1}\left(\frac{2}{5}\right)\)

\(\theta - 36.87^\circ = 23.58^\circ\)

\(\theta = 23.58^\circ + 36.87^\circ = 60.45^\circ\)

For the second solution:

\(\theta = 180^\circ - 23.58^\circ + 36.87^\circ = 193.29^\circ\)

(iii) The expression \(\frac{1}{4 \sin \theta - 3 \cos \theta + 6}\) has its greatest value when \(4 \sin \theta - 3 \cos \theta\) is minimized.

The minimum value of \(4 \sin \theta - 3 \cos \theta\) is \(-5\), so:

\(4 \sin \theta - 3 \cos \theta + 6 = 1\)

Thus, the greatest value of the expression is 1.

(a) Start with the equation:

\(\sqrt{5} \sec x + \tan x = 4\)

Express \(\sec x\) and \(\tan x\) in terms of \(\cos x\) and \(\sin x\):

\(\sqrt{5} \frac{1}{\cos x} + \frac{\sin x}{\cos x} = 4\)

Multiply through by \(\cos x\):

\(\sqrt{5} + \sin x = 4 \cos x\)

Rearrange to:

\(4 \cos x - \sin x = \sqrt{5}\)

Use the identity:

\(a \cos \theta \pm b \sin \theta \equiv R \cos(\theta \mp \alpha)\)

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

Here, \(a = 4\) and \(b = -1\):

\(R = \sqrt{4^2 + (-1)^2} = \sqrt{17}\)

\(\tan \alpha = \frac{-1}{4}\)

\(\alpha = \tan^{-1}\left(-\frac{1}{4}\right) \approx 14.04^\circ\)

(b) Solve \(\sqrt{5} \sec 2x + \tan 2x = 4\):

Using the result from (a):

\(4 \cos 2x - \sin 2x = \sqrt{5}\)

\(\cos(2x + 14.04^\circ) = \frac{\sqrt{5}}{\sqrt{17}}\)

\(2x + 14.04^\circ = \cos^{-1}\left(\frac{\sqrt{5}}{\sqrt{17}}\right)\)

Calculate \(\cos^{-1}\left(\frac{\sqrt{5}}{\sqrt{17}}\right) \approx 35.64^\circ\)

\(2x = 35.64^\circ - 14.04^\circ\)

\(2x = 21.6^\circ\)

\(x = 10.8^\circ\)

For the second solution:

\(2x = 180^\circ - 35.64^\circ + 14.04^\circ\)

\(2x = 158.4^\circ\)

\(x = 79.2^\circ\)

Thus, the solutions are \(x = 21.6^\circ\) and \(x = 144.4^\circ\).

(a) To express \(4 \cos x - \sin x\) in the form \(R \cos(x + \alpha)\), we use the identity:

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

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

Here, \(a = 4\) and \(b = 1\).

Calculate \(R\):

\(R = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{1}{4}\)

\(\alpha = \tan^{-1}\left(\frac{1}{4}\right) \approx 14.04^\circ\)

(b) Solve \(4 \cos 2x - \sin 2x = 3\) using the result from part (a):

\(R \cos(2x + \alpha) = 3\)

\(\sqrt{17} \cos(2x + 14.04^\circ) = 3\)

\(\cos(2x + 14.04^\circ) = \frac{3}{\sqrt{17}}\)

Calculate \(2x + 14.04^\circ\):

\(2x + 14.04^\circ = \cos^{-1}\left(\frac{3}{\sqrt{17}}\right)\)

\(2x + 14.04^\circ \approx 43.31^\circ\)

\(2x \approx 43.31^\circ - 14.04^\circ = 29.27^\circ\)

\(x \approx 14.6^\circ\)

For the second solution:

\(2x + 14.04^\circ = 360^\circ - 43.31^\circ\)

\(2x + 14.04^\circ \approx 316.69^\circ\)

\(2x \approx 316.69^\circ - 14.04^\circ = 302.65^\circ\)

\(x \approx 151.3^\circ\)