(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:

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

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

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

\(\theta = 117.3^\circ\)