Answer: amplitude \(4\), period \(120^\circ\); key points include maxima at \(-120^\circ,0^\circ,120^\circ\), minima at \(-60^\circ,60^\circ\), and \(\theta\)-intercepts \(\pm40^\circ,\pm80^\circ\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Use amplitude, period and vertical shift to identify the key points of the graph.

The function is

\(y=2+4\cos3\theta.\)

The amplitude is the coefficient of the cosine term, so the amplitude is

\(4.\)

The period of \(\cos3\theta\) is

\(\frac{360^\circ}{3}=120^\circ.\)

The midline is \(y=2\). Since the amplitude is \(4\), the maximum value is

\(2+4=6,\)

and the minimum value is

\(2-4=-2.\)

In the interval \(-120^\circ\leq\theta\leq120^\circ\), maxima occur when

\(3\theta=-360^\circ,0^\circ,360^\circ,\)

so

\(\theta=-120^\circ,0^\circ,120^\circ.\)

Minima occur when

\(3\theta=-180^\circ,180^\circ,\)

so

\(\theta=-60^\circ,60^\circ.\)

For the \(\theta\)-intercepts, set \(y=0\):

\(2+4\cos3\theta=0,\)

so

\(\cos3\theta=-\frac12.\)

This gives

\(\theta=-80^\circ,-40^\circ,40^\circ,80^\circ.\)

These points determine the required cosine sketch.

This completes the solution and gives the required result.