(i) To show that \(4x^2 - 1 = \cot x\) has only one root in the interval \(0 < x < \frac{1}{2}\pi\), sketch the graphs of \(y = 4x^2 - 1\) and \(y = \cot x\). The graph of \(y = 4x^2 - 1\) is a parabola opening upwards with vertex at \((0, -1)\). The graph of \(y = \cot x\) has vertical asymptotes at \(x = 0\) and \(x = \frac{1}{2}\pi\), and decreases from positive infinity to negative infinity. The two graphs intersect only once in the interval \(0 < x < \frac{1}{2}\pi\).

(ii) To verify the root lies between 0.6 and 1, calculate \(4x^2 - 1 - \cot x\) at \(x = 0.6\) and \(x = 1\). At \(x = 0.6\), \(4(0.6)^2 - 1 - \cot(0.6)\) is negative, and at \(x = 1\), \(4(1)^2 - 1 - \cot(1)\) is positive. This indicates a sign change, confirming a root exists between 0.6 and 1.

(iii) Using the iterative formula \(x_{n+1} = \frac{1}{2}\sqrt{1 + \cot x_n}\), start with an initial guess, say \(x_0 = 0.7\). Calculate successive iterations:

\(x_1 = \frac{1}{2}\sqrt{1 + \cot(0.7)}\)

\(x_2 = \frac{1}{2}\sqrt{1 + \cot(x_1)}\)

Continue until the value stabilizes to 2 decimal places. The iterations converge to approximately 0.73.