(a) The length of CD is given by CD = 2r - 2r \cos x. The area of the trapezium is \(\frac{1}{2} \times (AB + CD) \times r \sin x = \frac{1}{2} \times (2r + 2r - 2r \cos x) \times r \sin x = r^2 (2 - \cos x) \sin x\).

The area of each sector is \(\frac{1}{2} r^2 x\), so the total area of the sectors is \(r^2 x\). Given that this is 90% of the area of the trapezium, we have \(r^2 x = 0.9 \times r^2 (2 - \cos x) \sin x\).

Canceling \(r^2\) from both sides gives \(x = 0.9 (2 - \cos x) \sin x\).

(b) Calculate \(0.9 (2 - \cos 0.5) \sin 0.5 \approx 0.484\) and \(0.9 (2 - \cos 0.7) \sin 0.7 \approx 0.716\). Since \(0.5 < 0.484 < 0.7\), x lies between 0.5 and 0.7.

(c) The iterative formula \(x_{n+1} = \cos^{-1} \left( \frac{2 - x_n}{0.9 \sin x_n} \right)\) rearranges to \(x = 0.9 \sin x (2 - \cos x)\), which is the equation from part (a). Therefore, if the sequence converges, it converges to the root of the equation.

\((d) Using the iterative formula, starting with an initial guess of x = 0.6:\)

1. \(x_1 = \cos^{-1} \left( \frac{2 - 0.6}{0.9 \sin 0.6} \right) \approx 0.6205\)

2. \(x_2 = \cos^{-1} \left( \frac{2 - 0.6205}{0.9 \sin 0.6205} \right) \approx 0.6199\)

3. \(x_3 = \cos^{-1} \left( \frac{2 - 0.6199}{0.9 \sin 0.6199} \right) \approx 0.6200\)

The value converges to 0.62 correct to 2 decimal places.