\(3\cos \theta + 4\sin \theta\) may be written in the form \(r\cos(\theta - \alpha)\), where \(r > 0\).

- Calculate the value of \(r\), and show that one value of \(\alpha\) is approximately \(53.1^\circ\).

We have the compound angle identity, \[\cos (\theta -\alpha) = \cos\theta\cos\alpha+\sin\theta\sin\alpha.\]

Thus \[r\cos (\theta -\alpha) = (r\cos\alpha)\cos\theta+(r\sin\alpha)\sin\theta.\]

Comparing this with \(3\cos \theta + 4\sin \theta\) gives us that \[r\cos\alpha = 3, r\sin\alpha = 4.\]

Squaring and adding these two equations gives us that \(r^2 = 25\), so \(r = 5\).

Dividing the two equations gives us that \(\tan \alpha = \dfrac{4}{3}\), and one solution of this is \(\alpha \approx 53.1^\circ\).

- Hence show that one solution of \(3\cos \theta + 4\sin \theta = 2\) is approximately \(120^\circ\).

Using the results above, we can write the equation as \(5\cos(\theta - 53.1^\circ) = 2.\) Two of the solutions to this are \[\theta - 53.1^\circ = \cos^{-1} 0.4 \quad\implies\quad \theta \approx 119.5^\circ\] and \[\theta - 53.1^\circ = -\cos^{-1} 0.4 \quad\implies\quad \theta \approx -13.3^\circ.\]

Thus \(120^\circ\) is an approximate solution to the equation.

- State all the other solutions for \(0^\circ \leq \theta \leq 500^\circ.\)

As well as the above solutions, we can find others by adding integer multiples of \(360^\circ\). We find that \(119.5^\circ\), \(346.7^\circ\) and \(479.5^\circ\) are all the solutions within the given range.

- Hence give the positive values of \(\theta\) less than \(500^\circ\) for which \(3\cos \theta + 4\sin \theta > 2\).

A graph of the function \(y=3\cos \theta + 4\sin \theta\) looks like this.

Note that the function switches between being above and below \(y=2\) at each of the intersections found in part (c). When \(\theta = 0\), the function is greater than \(2\).

This must mean that \(3\cos \theta + 4\sin \theta > 2\) on the intervals \(0 < \theta < 119.5^\circ\) and \(346.7^\circ < \theta < 479.5^\circ\).