Answer: \(a=4\), \(b=\frac38\), \(c=2\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

The period of \(a\cos bx+c\), with \(x\) in degrees, is \(\frac{360^\circ}{b}\).

Since the period is \(960^\circ\),

\(\frac{360}{b}=960\), so \(b=\frac38\).

From the graph, \(\mathrm f(0)=6\). Since \(\cos0=1\),

\(a+c=6\).

Also \(\mathrm f(320^\circ)=0\). Now \(\frac38\cdot320^\circ=120^\circ\), and \(\cos120^\circ=-\frac12\), so

\(-\frac{a}{2}+c=0\).

Thus \(c=\frac{a}{2}\). Substitute into \(a+c=6\):

\(a+\frac{a}{2}=6\), so \(a=4\), and \(c=2\).

This gives the required answer: \(a=4\), \(b=\frac38\), \(c=2\).