(a) The area of the major sector is \(\frac{1}{2} r^2 (2\pi - x)\).

The area of the shaded segment is \(\frac{1}{2} r^2 x - \frac{1}{2} r^2 \sin x\).

Given that the area of the major sector is 3 times the area of the shaded region:

\(\frac{1}{2} r^2 (2\pi - x) = 3 \left( \frac{1}{2} r^2 x - \frac{1}{2} r^2 \sin x \right)\)

Simplifying gives:

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

\(2\pi = 4x - 3\sin x\)

\(x = \frac{3}{4} \sin x + \frac{1}{2} \pi\)

(b) Calculate the values at \(x = 2\) and \(x = 2.5\):

For \(x = 2\), \(\frac{3}{4} \sin 2 + \frac{1}{2} \pi \approx 2.2528\)

For \(x = 2.5\), \(\frac{3}{4} \sin 2.5 + \frac{1}{2} \pi \approx 2.0197\)

Since \(2.2528 > 2\) and \(2.0197 < 2.5\), the root lies between 2 and 2.5.

(c) Using the iterative formula:

\(x_{n+1} = \frac{3}{4} \sin x_n + \frac{1}{2} \pi\)

Starting with \(x_1 = 2\):

\(x_2 = \frac{3}{4} \sin 2 + \frac{1}{2} \pi \approx 2.2528\)

\(x_3 = \frac{3}{4} \sin 2.2528 + \frac{1}{2} \pi \approx 2.1530\)

\(x_4 = \frac{3}{4} \sin 2.1530 + \frac{1}{2} \pi \approx 2.1972\)

\(x_5 = \frac{3}{4} \sin 2.1972 + \frac{1}{2} \pi \approx 2.1784\)

\(x_6 = \frac{3}{4} \sin 2.1784 + \frac{1}{2} \pi \approx 2.1866\)

\(x_7 = \frac{3}{4} \sin 2.1866 + \frac{1}{2} \pi \approx 2.1830\)

\(x_8 = \frac{3}{4} \sin 2.1830 + \frac{1}{2} \pi \approx 2.1846\)

The root is approximately 2.18 to 2 decimal places.