(i) The graph of \(y = \sin x\) starts at \((0,0)\), peaks at \((90^\circ, 1)\), and returns to \((180^\circ, 0)\). The graph of \(y = \cos 2x\) starts at \((0, 1)\), completes one full cycle by \(180^\circ\), and oscillates between -1 and 1.

(ii) To verify \(x = 30^\circ\) is a root, calculate \(\sin 30^\circ = 0.5\) and \(\cos 60^\circ = 0.5\), confirming \(\sin 30^\circ = \cos 60^\circ\). The other root is found by solving \(\sin x = \cos 2x\) for \(0^\circ \leq x \leq 180^\circ\), giving \(x = 150^\circ\).

(iii) The inequality \(\sin x < \cos 2x\) holds for \(0 \leq x < 30\) and \(150 < x \leq 180\).

(i) The graph of \(y = \cos 2\theta\) oscillates between -1 and 1 and completes two full oscillations in the interval \([0, 2\pi]\). The line \(y = \frac{1}{2}\) is a horizontal line. The intersections of these two graphs represent the solutions to the equation \(2\cos 2\theta - 1 = 0\).

(ii) The equation \(2\cos 2\theta - 1 = 0\) simplifies to \(\cos 2\theta = \frac{1}{2}\). The solutions for \(2\theta\) are \(60^\circ, 300^\circ\) (or \(\frac{\pi}{3}, \frac{5\pi}{3}\)) within one period \([0, 2\pi]\). Since \(2\theta\) ranges from \(0\) to \(4\pi\), there are 4 solutions: \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

(iii) The interval \(10\pi \leq \theta \leq 20\pi\) is 5 times the length of \(0 \leq \theta \leq 2\pi\). Therefore, the number of roots is 5 times the number of roots in the interval \(0 \leq \theta \leq 2\pi\), which is \(5 \times 4 = 20\).

(i) The curve \(y = 2 \sin x\) is a sine wave with amplitude 2, period \(2\pi\), and it oscillates between -2 and 2. The curve completes one full cycle from \(0\) to \(2\pi\).

(ii) To find the number of real roots of the equation \(2\pi \sin x = \pi - x\), we rearrange it to \(2\pi \sin x + x = \pi\). This can be rewritten as \(y = 2\pi \sin x\) and \(y = \pi - x\). The line \(y = \pi - x\) can be rewritten as \(y = 1 - \frac{x}{\pi}\) by dividing through by \(\pi\).

The line \(y = 1 - \frac{x}{\pi}\) passes through the points \((0, 1)\) and \((\pi, 0)\). By sketching both the sine curve and the line, we observe that they intersect at three points within the interval \(0 \leq x \leq 2\pi\), indicating there are 3 real roots.

The curve \(y = 3 \cos 2x\) is a cosine function with amplitude 3 and period \(\pi\). It completes one full oscillation from \(x = 0\) to \(x = \pi\). The range of the curve is from -3 to 3.

The line \(x + 2y = \pi\) can be rewritten as \(y = \frac{\pi - x}{2}\). This is a straight line with a negative slope of -1/2 and a y-intercept of \(\frac{\pi}{2}\).

To sketch the curve, plot the cosine wave starting at \(y = 3\) when \(x = 0\), decreasing to \(y = -3\) at \(x = \frac{\pi}{2}\), and returning to \(y = 3\) at \(x = \pi\).

To sketch the line, plot the intercepts: when \(x = 0\), \(y = \frac{\pi}{2}\); when \(y = 0\), \(x = \pi\). Draw the line through these points.

(i) Given the maximum value of \(f(x) = a - b \cos x\) is 10, and the minimum value is \(-2\), we have:

\(a + b = 10\) (when \(\cos x = -1\))

\(a - b = -2\) (when \(\cos x = 1\))

Solving these equations simultaneously:

\(a + b = 10\)

\(a - b = -2\)

Adding the equations gives:

\(2a = 8 \Rightarrow a = 4\)

Substituting \(a = 4\) into \(a + b = 10\):

\(4 + b = 10 \Rightarrow b = 6\)

(ii) To solve \(f(x) = 0\):

\(4 - 6 \cos x = 0\)

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

Using the inverse cosine function, \(x = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ\) or \(360^\circ - 48.2^\circ = 311.8^\circ\).

(iii) The sketch of \(y = f(x)\) should show one cycle of a cosine wave, starting at \(-2\), peaking at 10, and returning to \(-2\) at \(x = 360^\circ\).