(i) The graph of \(y = 3 \sin x\) is a sine wave with amplitude 3, oscillating between -3 and 3, over the interval \(-\pi \leq x \leq \pi\). The maximum point is at \(x = \frac{\pi}{2}\), where \(y = 3\).

(ii) The line \(y = kx\) passes through the maximum point \(\left( \frac{\pi}{2}, 3 \right)\). Substituting into the line equation gives:

\(3 = k \left( \frac{\pi}{2} \right)\)

Solving for \(k\):

\(k = \frac{6}{\pi}\)

(iii) To find the other intersection point, set \(y = 3 \sin x = kx\) and solve for \(x\):

\(3 \sin x = \frac{6}{\pi} x\)

At \(x = -\frac{\pi}{2}\), \(\sin x = -1\), so:

\(3(-1) = \frac{6}{\pi} \left(-\frac{\pi}{2}\right)\)

\(-3 = -3\)

Thus, the coordinates are \(\left( -\frac{\pi}{2}, -3 \right)\).

(i) We have two equations from the given conditions:

\(a \sin\left(\frac{1}{2}\pi\right) + b = 2\)

\(a \cdot 1 + b = 2\)

\(a + b = 2\)

\(a \sin\left(\frac{3}{2}\pi\right) + b = -8\)

\(a \cdot (-1) + b = -8\)

\(-a + b = -8\)

Solving these equations:

\(a + b = 2\)

\(-a + b = -8\)

Add the equations:

\(2b = -6\)

\(b = -3\)

Substitute \(b = -3\) into \(a + b = 2\):

\(a - 3 = 2\)

\(a = 5\)

(ii) To find \(x\) for which \(f(x) = 0\):

\(5 \sin x - 3 = 0\)

\(5 \sin x = 3\)

\(\sin x = \frac{3}{5}\)

\(x = \sin^{-1}\left(\frac{3}{5}\right) \approx 0.64\)

Since \(\sin x\) is positive in the first and second quadrants:

\(x = \pi - 0.64 \approx 2.50\)

(iii) The graph of \(y = 5 \sin x - 3\) is a sine wave with amplitude 5, shifted down by 3 units. It starts at \(y = -3\) when \(x = 0\), reaches a maximum of \(y = 2\) at \(x = \frac{1}{2}\pi\), and a minimum of \(y = -8\) at \(x = \frac{3}{2}\pi\).

(i) To solve \(fg(x) = 1\), substitute \(g(x) = \frac{1}{2}x\) into \(f(x)\):

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

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

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

\(x = 2 \times \cos^{-1}\left(\frac{1}{3}\right) \approx 2.46\)

Alternative: Solve \(f(y) = 1 \rightarrow y = 1.23 \rightarrow \frac{1}{2}x = 1.23 \rightarrow x = 2.46\)

(ii) The graph of \(y = f(x) = 2 - 3 \cos x\) is a cosine curve. Sketch one cycle from \(0\) to \(2\pi\), starting and ending at the same negative \(y\) value, with a peak at \(y = 5\) and a trough at \(y = -1\).

(i) Start with the equation \(2 \cos x + 3 \sin x = 0\).

Rearrange to \(\tan x = -\frac{2}{3}\).

Find \(x\) using the inverse tangent function: \(x = 146.3^\circ\) or \(x = 326.3^\circ\).

(ii) Sketch the graphs of \(y = 2 \cos x\) and \(y = -3 \sin x\) over the interval \(0^\circ \leq x \leq 360^\circ\). The cosine graph starts at the top of the curve, while the sine graph is inverted and starts on the x-axis.

(iii) From the graphs, determine where \(2 \cos x + 3 \sin x > 0\). This occurs when \(x < 146.3^\circ\) and \(x > 326.3^\circ\).

(i) The curve \(y = \sin 2x\) completes two cycles from \(0\) to \(2\pi\), starting and finishing on the x-axis, with a maximum of 1 and a minimum of -1. The curve \(y = \cos x - 1\) completes one cycle, starting and finishing on the x-axis, with a minimum point at -2.

(ii) (a) For \(2 \sin 2x + 1 = 0\), we have \(\sin 2x = -\frac{1}{2}\). The solutions for \(\sin 2x = -\frac{1}{2}\) in the interval \(0 \leq x \leq 2\pi\) are four points where the sine curve intersects the line \(y = -\frac{1}{2}\).

(b) For \(\sin 2x - \cos x + 1 = 0\), the solutions are the points where the curves \(y = \sin 2x\) and \(y = \cos x - 1\) intersect. There are three such intersections in the interval \(0 \leq x \leq 2\pi\).