(i) Sketch the graphs of \(y = 2 \cot x\) and \(y = 1 + e^x\). The intersection of these graphs in the interval \(0 < x < \frac{1}{2} \pi\) indicates the root. The behavior of \(2 \cot x\) and \(1 + e^x\) suggests only one intersection point in this interval.

(ii) Evaluate \(2 \cot x - 1 - e^x\) at \(x = 0.5\) and \(x = 1.0\). The sign change between these points confirms a root exists between them.

(iii) Rearrange the original equation to show it is equivalent to \(x = \arctan\left(\frac{2}{1 + e^x}\right)\).

(iv) Use the iterative formula:

\(x_1 = 0.7\)

\(x_2 = \arctan\left(\frac{2}{1 + e^{0.7}}\right) = 0.6104\)

\(x_3 = \arctan\left(\frac{2}{1 + e^{0.6104}}\right) = 0.6100\)

\(x_4 = \arctan\left(\frac{2}{1 + e^{0.6100}}\right) = 0.6100\)

The root correct to 2 decimal places is 0.61.