(i) The length of the tangent CT can be expressed as \(CT = r \tan x\). The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the sector BOC is \(\frac{1}{2} r^2 x\). The area of the triangle OCT is \(\frac{1}{2} r \cdot r \tan x = \frac{1}{2} r^2 \tan x\). The area of the shaded region is the area of the triangle minus the area of the sector: \(\frac{1}{2} r^2 \tan x - \frac{1}{2} r^2 x\). Setting this equal to the area of the semicircle gives:

\(\frac{1}{2} r^2 \tan x - \frac{1}{2} r^2 x = \frac{1}{2} \pi r^2\)

Dividing through by \(\frac{1}{2} r^2\) gives:

\(\tan x - x = \pi\)

Thus, \(\tan x = x + \pi\).

(ii) Using the iterative formula \(x_{n+1} = \arctan(x_n + \pi)\), start with an initial guess, say \(x_0 = 1\).

1. \(x_1 = \arctan(1 + \pi) = 1.3521\)

2. \(x_2 = \arctan(1.3521 + \pi) = 1.3490\)

3. \(x_3 = \arctan(1.3490 + \pi) = 1.3492\)

4. \(x_4 = \arctan(1.3492 + \pi) = 1.3492\)

The value converges to 1.35 when rounded to 2 decimal places.