Answer: \(a=6\), \(b=2\), \(c=-3\); \(P=\left(\frac{\pi}{6},0\right)\); \(Q=\left(\frac{\pi}{2},-9\right)\).

The period of \(a\cos bx+c\) is

\(\frac{2\pi}{b}.\)

Since the period is \(\pi\),

\(\frac{2\pi}{b}=\pi,\)

so

\(b=2.\)

The graph passes through \((0,3)\), so

\(3=a\cos0+c=a+c. \tag{1}\)

It also passes through \(R\left(\frac{5\pi}{6},0\right)\), so

\(0=a\cos\left(2\cdot\frac{5\pi}{6}\right)+c.\)

Thus

\(0=a\cos\frac{5\pi}{3}+c=\frac a2+c. \tag{2}\)

Solving (1) and (2),

\(a=6,\qquad c=-3.\)

Therefore

\(y=6\cos2x-3.\)

(b) For the \(x\)-intercepts,

\(6\cos2x-3=0.\)

So

\(\cos2x=\frac12.\)

In the given interval, the intercepts are at

Since \(R\) is the intercept at \(\frac{5\pi}{6}\),

\(P=\left(\frac{\pi}{6},0\right).\)

(c) The minimum occurs when \(\cos2x=-1\). Hence

\(2x=\pi,\)

so

\(x=\frac{\pi}{2}.\)

The corresponding \(y\)-value is

\(y=6(-1)-3=-9.\)

Therefore

\(Q=\left(\frac{\pi}{2},-9\right).\)