The graph of \(y = a + b \sin x\) is a vertically shifted sine wave.

The midline of the sine wave is at \(y = 1\), which indicates that \(a = 1\).

The amplitude of the sine wave is the distance from the midline to the maximum or minimum point. The maximum value is \(3\) and the minimum value is \(-1\), so the amplitude is \(\frac{3 - 1}{2} = 2\).

Thus, \(b = 2\).

(i) The amplitude of the sine function is the distance from the midline to the maximum, which is \(9 - 3 = 6\). Therefore, \(a = 6\).

The period of the sine function is \(2\pi\) divided by the coefficient of \(x\), which is \(b\). The period is \(\pi\), so \(b = 2\).

The vertical shift \(c\) is the midline of the graph, which is at \(y = 3\). Therefore, \(c = 3\).

(ii) To find the smallest \(x\) for which \(y = 0\), set the equation to zero: \(6\sin(2x) + 3 = 0\).

Solving for \(\sin(2x)\):

\(6\sin(2x) = -3\)

\(\sin(2x) = -\frac{1}{2}\)

The general solution for \(\sin(\theta) = -\frac{1}{2}\) is \(\theta = \frac{7\pi}{6} + 2k\pi\) or \(\theta = \frac{11\pi}{6} + 2k\pi\).

For \(2x = \frac{7\pi}{6}\), solve for \(x\):

\(x = \frac{7\pi}{12}\)

Converting to decimal gives \(x \approx 1.83\).

(a) The function \(y = 3 \cos 2x + 2\) is a cosine function scaled and shifted. The cosine function \(\cos 2x\) has a range of \([-1, 1]\). Therefore, \(3 \cos 2x\) has a range of \([-3, 3]\). Adding 2 shifts this range to \([-1, 5]\). Thus, the greatest value is 5 and the least value is -1.

(b) The graph of \(y = 3 \cos 2x + 2\) is a cosine wave starting at its maximum value of 5 at \(x = 0\), decreasing to its minimum value of -1 at \(x = \frac{\pi}{2}\), and returning to its maximum value of 5 at \(x = \pi\). This completes one full cycle.

(c)(i) For \(k = -3\), the line \(y = -3x\) does not intersect the curve \(y = 3 \cos 2x + 2\) within the given range, so there are 0 solutions.

(c)(ii) For \(k = 1\), the line \(y = x\) intersects the curve \(y = 3 \cos 2x + 2\) at two points within the given range, so there are 2 solutions.

(c)(iii) For \(k = 3\), the line \(y = 3x\) intersects the curve \(y = 3 \cos 2x + 2\) at one point within the given range, so there is 1 solution.

(i) The function is given by \(f(x) = a + b \cos 2x\). We have two conditions:

\(f(0) = a + b \cos 0 = -1\) which simplifies to \(a + b = -1\).

\(f\left(\frac{1}{2}\pi\right) = a + b \cos \pi = 7\) which simplifies to \(a - b = 7\).

Solving these two equations:

\(a + b = -1\)

\(a - b = 7\)

Add the equations: \(2a = 6\) so \(a = 3\).

Substitute \(a = 3\) into \(a + b = -1\):

\(3 + b = -1\) so \(b = -4\).

Thus, \(a = 3\) and \(b = -4\).

(ii) To find the x-coordinates where the curve intersects the x-axis, set \(f(x) = 0\):

\(3 - 4 \cos 2x = 0\)

\(\cos 2x = \frac{3}{4}\)

Solving for \(x\), we find:

\(2x = \cos^{-1}\left(\frac{3}{4}\right)\)

\(x = \frac{1}{2} \cos^{-1}\left(\frac{3}{4}\right)\)

Calculating gives \(x \approx 0.36\) and \(x \approx 2.78\).

(iii) The graph of \(y = f(x)\) is a cosine wave starting at \(y = -1\) when \(x = 0\) and reaching a maximum of \(y = 7\) at \(x = \frac{1}{2}\pi\), then returning to \(y = -1\) at \(x = \pi\). The graph oscillates once between \(x = 0\) and \(x = \pi\).

(i) To sketch the graphs of \(y = 2 \sin x\) and \(y = \cos 2x\) for \(0 \leq x \leq \pi\):

- The graph of \(y = 2 \sin x\) will have a maximum value of 2 and a minimum value of 0 within the interval. It completes half a cycle from 0 to \(\pi\).

- The graph of \(y = \cos 2x\) completes a full cycle from 0 to \(\pi\), oscillating between 1 and -1.

(ii) The number of solutions to \(2 \sin x = \cos 2x\) corresponds to the number of intersection points of the two graphs within the interval \(0 \leq x \leq \pi\). From the sketch, there are 2 points of intersection.