(a) To express \(3 \cos x + 2 \cos(x - 60^\circ)\) in the form \(R \cos(x - \alpha)\):

First, expand \(2 \cos(x - 60^\circ)\):

\(2 \cos(x - 60^\circ) = 2(\cos x \cos 60^\circ + \sin x \sin 60^\circ) = \cos x + \sqrt{3} \sin x\)

Combine with \(3 \cos x\):

\(3 \cos x + \cos x + \sqrt{3} \sin x = 4 \cos x + \sqrt{3} \sin x\)

Now, 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 \(b = \sqrt{3}\):

\(R = \sqrt{4^2 + (\sqrt{3})^2} = \sqrt{16 + 3} = \sqrt{19}\) \(\tan \alpha = \frac{\sqrt{3}}{4}\)

Thus, \(\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \approx 23.41^\circ\).

(b) Solve \(3 \cos 2\theta + 2 \cos(2\theta - 60^\circ) = 2.5\):

Using the result from part (a), express:

\(3 \cos 2\theta + 2 \cos(2\theta - 60^\circ) = \sqrt{19} \cos(2\theta - 23.41^\circ)\)

Set equal to 2.5:

\(\sqrt{19} \cos(2\theta - 23.41^\circ) = 2.5\)

\(\cos(2\theta - 23.41^\circ) = \frac{2.5}{\sqrt{19}}\)

Find \(2\theta\):

\(2\theta - 23.41^\circ = \cos^{-1}\left(\frac{2.5}{\sqrt{19}}\right)\)

Calculate \(2\theta\):

\(2\theta = \cos^{-1}\left(\frac{2.5}{\sqrt{19}}\right) + 23.41^\circ\)

\(\theta = 39.2^\circ\) and \(\theta = 82.1^\circ\) (considering the range \(0^\circ < \theta < 180^\circ\)).

(i) To express \(\sqrt{6} \sin x + \cos x\) 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, \(a = \sqrt{6}\) and \(b = 1\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{6})^2 + 1^2} = \sqrt{6 + 1} = \sqrt{7}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{1}{\sqrt{6}}\)

\(\alpha = \tan^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 22.208^\circ\)

(ii) Solve \(\sqrt{6} \sin 2\theta + \cos 2\theta = 2\).

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

\(\sqrt{6} \sin 2\theta + \cos 2\theta = \sqrt{7} \sin(2\theta + 22.208^\circ)\)

Set equal to 2:

\(\sqrt{7} \sin(2\theta + 22.208^\circ) = 2\)

\(\sin(2\theta + 22.208^\circ) = \frac{2}{\sqrt{7}}\)

Calculate \(2\theta + 22.208^\circ\):

\(2\theta + 22.208^\circ = \sin^{-1}\left(\frac{2}{\sqrt{7}}\right) \approx 49.107^\circ\)

\(2\theta = 49.107^\circ - 22.208^\circ\)

\(2\theta = 26.899^\circ\)

\(\theta = 13.4495^\circ \approx 13.4^\circ\)

For the second solution:

\(2\theta + 22.208^\circ = 180^\circ - 49.107^\circ\)

\(2\theta = 180^\circ - 49.107^\circ - 22.208^\circ\)

\(2\theta = 108.685^\circ\)

\(\theta = 54.3425^\circ \approx 54.3^\circ\)

(i) Start with the equation:

\(\sqrt{2} \csc x + \cot x = \sqrt{3}\)

Express \(\csc x\) and \(\cot x\) in terms of sine and cosine:

\(\sqrt{2} \left( \frac{1}{\sin x} \right) + \frac{\cos x}{\sin x} = \sqrt{3}\)

Combine into a single fraction:

\(\frac{\sqrt{2} + \cos x}{\sin x} = \sqrt{3}\)

Multiply through by \(\sin x\):

\(\sqrt{2} + \cos x = \sqrt{3} \sin x\)

Rearrange to:

\(\sqrt{3} \sin x - \cos x = \sqrt{2}\)

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, \(a = \sqrt{3}\) and \(b = 1\).

Calculate \(R\):

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

Calculate \(\alpha\):

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

\(\alpha = 30^\circ\)

Thus, \(\sqrt{3} \sin x - \cos x = 2 \sin(x - 30^\circ)\).

(ii) Solve \(2 \sin(x - 30^\circ) = \sqrt{2}\):

\(\sin(x - 30^\circ) = \frac{\sqrt{2}}{2}\)

\(x - 30^\circ = 45^\circ\) or \(x - 30^\circ = 135^\circ\)

\(x = 75^\circ\) or \(x = 165^\circ\)

(i) Expand \(2 \sin(x - 30^\circ)\) using the angle subtraction formula:

\(2 \sin(x - 30^\circ) = 2(\sin x \cos 30^\circ - \cos x \sin 30^\circ)\)

\(= \sqrt{3} \sin x - \cos x\)

Combine terms: \(\sqrt{3} \sin x - 2 \cos x\).

Using the identity:

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

\(R = \sqrt{a^2 + b^2} = \sqrt{(\sqrt{3})^2 + (-2)^2} = \sqrt{7}\)

\(\tan \alpha = \frac{b}{a} = \frac{-2}{\sqrt{3}}\)

\(\alpha = \tan^{-1}\left(\frac{-2}{\sqrt{3}}\right) \approx 49.11^\circ\)

(ii) Solve \(\sqrt{3} \sin x - 2 \cos x = 1\):

\(R \sin(x - \alpha) = 1\)

\(\sqrt{7} \sin(x - 49.11^\circ) = 1\)

\(\sin(x - 49.11^\circ) = \frac{1}{\sqrt{7}}\)

\(x - 49.11^\circ = \sin^{-1}\left(\frac{1}{\sqrt{7}}\right) \approx 22.21^\circ\)

\(x = 22.21^\circ + 49.11^\circ = 71.3^\circ\)

(i) To express \(8 \cos \theta - 15 \sin \theta\) in the form \(R \cos(\theta + \alpha)\), 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, \(a = 8\) 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 \(8 \cos 2x - 15 \sin 2x = 4\).

Using the result from part (i), express the equation as:

\(17 \cos(2x + 61.93^\circ) = 4\)

\(\cos(2x + 61.93^\circ) = \frac{4}{17}\)

\(2x + 61.93^\circ = \cos^{-1}\left(\frac{4}{17}\right) \approx 76.39^\circ\)

\(2x = 76.39^\circ - 61.93^\circ = 14.46^\circ\)

\(x = \frac{14.46^\circ}{2} = 7.23^\circ\)

For the second solution:

\(2x + 61.93^\circ = 360^\circ - 76.39^\circ = 283.61^\circ\)

\(2x = 283.61^\circ - 61.93^\circ = 221.68^\circ\)

\(x = \frac{221.68^\circ}{2} = 110.84^\circ\)

Thus, the solutions are \(x = 7.2^\circ\) and \(x = 110.8^\circ\).

(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:

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

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

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

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

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

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

(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 \(\theta \approx 27.4^\circ, 117.4^\circ\).

(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:

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

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

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

\(x \approx 151.3^\circ\)

(a) Expand \(\cos(x - 60^\circ)\) using the formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\):

\(\cos(x - 60^\circ) = \cos x \cos 60^\circ + \sin x \sin 60^\circ\)

\(= \cos x \cdot \frac{1}{2} + \sin x \cdot \frac{\sqrt{3}}{2}\)

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

Substitute into \(2\cos(x - 60^\circ) + \cos x\):

\(2\left(\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x\right) + \cos x\)

\(= \cos x + \sqrt{3} \sin x + \cos x\)

\(= 2\cos x + \sqrt{3} \sin x\)

Now, express in the form \(R\cos(x - \alpha)\):

\(R = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}\)

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

\(\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \approx 40.89^\circ\)

(b) The expression \(R\cos(x - \alpha)\) takes its least value when \(\cos(x - \alpha) = -1\):

\(x - \alpha = 180^\circ\)

\(x = 180^\circ + \alpha\)

\(x = 180^\circ + 40.89^\circ = 220.89^\circ\)

(a) To express \(5 \sin x - 3 \cos x\) in the form \(R \sin(x - \alpha)\), 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, \(a = 5\) and \(b = -3\).

Calculate \(R\):

\(R = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-3}{5}\)

\(\alpha = \arctan\left(\frac{-3}{5}\right)\)

\(\alpha \approx 0.54\) radians (to 2 decimal places).

(b) The maximum value of \(R \sin(x - \alpha)\) is \(R\) and the minimum value is \(-R\).

Thus, the greatest value of \((5 \sin x - 3 \cos x)^2\) is:

\(R^2 = (\sqrt{34})^2 = 34\)

The least value is:

\(0\)

**(a)** To express \(\sqrt{7} \sin x + 2 \cos x\) in the form \(R \sin(x + \alpha)\), 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, \(a = \sqrt{7}\) and \(b = 2\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{7})^2 + 2^2} = \sqrt{7 + 4} = \sqrt{11}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{2}{\sqrt{7}}\)

\(\alpha = \tan^{-1}\left(\frac{2}{\sqrt{7}}\right) \approx 37.09^\circ\)

**(b)** Solve \(\sqrt{7} \sin 2\theta + 2 \cos 2\theta = 1\).

Using the result from part (a), express the equation as:

\(\sqrt{11} \sin(2\theta + 37.09^\circ) = 1\)

\(\sin(2\theta + 37.09^\circ) = \frac{1}{\sqrt{11}}\)

Find the angle:

\(2\theta + 37.09^\circ = \sin^{-1}\left(\frac{1}{\sqrt{11}}\right) \approx 17.5484^\circ\)

Thus, \(2\theta = 17.5484^\circ - 37.09^\circ\) or \(2\theta = 180^\circ - 17.5484^\circ - 37.09^\circ\).

Calculate \(\theta\):

\(\theta = \frac{17.5484^\circ - 37.09^\circ}{2} \approx 62.7^\circ\)

\(\theta = \frac{180^\circ - 17.5484^\circ - 37.09^\circ}{2} \approx 170.2^\circ\)

(a) We need to express \(\sqrt{6} \cos \theta + 3 \sin \theta\) in the form \(R \cos(\theta - \alpha)\).

Using 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, \(a = \sqrt{6}\) and \(b = 3\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{6})^2 + 3^2} = \sqrt{6 + 9} = \sqrt{15}\)

Calculate \(\alpha\):

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

\(\alpha = \tan^{-1}\left(\frac{3}{\sqrt{6}}\right) \approx 50.77^\circ\)

(b) Solve \(\sqrt{6} \cos \frac{1}{3}x + 3 \sin \frac{1}{3}x = 2.5\).

Using the result from part (a), we have:

\(\sqrt{15} \cos\left(\frac{1}{3}x - 50.77^\circ\right) = 2.5\)

\(\cos\left(\frac{1}{3}x - 50.77^\circ\right) = \frac{2.5}{\sqrt{15}}\)

Let \(\beta = \cos^{-1}\left(\frac{2.5}{\sqrt{15}}\right) \approx 49.797^\circ\).

Then:

\(\frac{1}{3}x - 50.77^\circ = 49.797^\circ\)

\(\frac{1}{3}x = 49.797^\circ + 50.77^\circ\)

\(x = 3(49.797^\circ + 50.77^\circ) \approx 301.7^\circ\)

For the second solution:

\(\frac{1}{3}x - 50.77^\circ = -49.797^\circ\)

\(\frac{1}{3}x = -49.797^\circ + 50.77^\circ\)

\(x = 3(-49.797^\circ + 50.77^\circ) \approx 2.9^\circ\)

(a) To express \(\sqrt{2} \cos x - \sqrt{5} \sin x\) 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, \(a = \sqrt{2}\) and \(b = -\sqrt{5}\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{2})^2 + (-\sqrt{5})^2} = \sqrt{2 + 5} = \sqrt{7}\)

Calculate \(\alpha\):

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

\(\alpha = \tan^{-1}\left(\frac{-\sqrt{5}}{\sqrt{2}}\right) \approx 57.688^\circ\)

(b) Solve \(\sqrt{2} \cos 2\theta - \sqrt{5} \sin 2\theta = 1\).

Using the result from part (a), express the equation as:

\(\sqrt{7} \cos(2\theta + 57.688^\circ) = 1\)

\(\cos(2\theta + 57.688^\circ) = \frac{1}{\sqrt{7}}\)

Find \(2\theta + 57.688^\circ\):

\(2\theta + 57.688^\circ = \cos^{-1}\left(\frac{1}{\sqrt{7}}\right) \approx 67.792^\circ\)

\(2\theta = 67.792^\circ - 57.688^\circ\)

\(2\theta = 10.104^\circ\)

\(\theta = 5.1^\circ\)

For the second solution:

\(2\theta + 57.688^\circ = 360^\circ - 67.792^\circ\)

\(2\theta = 292.208^\circ - 57.688^\circ\)

\(2\theta = 234.52^\circ\)

\(\theta = 117.3^\circ\)