(a) Sketch the graphs of \(y = \cot x\) and \(y = 2 - \cos x\) for \(0 < x \leq \frac{1}{2}\pi\). The intersection point of these graphs indicates the root of the equation \(\cot x = 2 - \cos x\). The graph shows one intersection in the given interval.

(b) Calculate \(\cot(0.6)\) and \(\cot(0.8)\) and compare with \(2 - \cos(0.6)\) and \(2 - \cos(0.8)\). For \(x = 0.6\), \(\cot(0.6) \approx 1.46\) and \(2 - \cos(0.6) \approx 1.34\). For \(x = 0.8\), \(\cot(0.8) \approx 1.03\) and \(2 - \cos(0.8) \approx 1.29\). The change in sign between these values indicates a root between 0.6 and 0.8.

(c) Use the iterative formula \(x_{n+1} = \arctan\left( \frac{1}{2 - \cos x_n} \right)\) starting with an initial guess, such as \(x_0 = 0.7\). Perform iterations:

\(x_1 = \arctan\left( \frac{1}{2 - \cos(0.7)} \right) \approx 0.6806\)

\(x_2 = \arctan\left( \frac{1}{2 - \cos(0.6806)} \right) \approx 0.6855\)

\(x_3 = \arctan\left( \frac{1}{2 - \cos(0.6855)} \right) \approx 0.6843\)

\(x_4 = \arctan\left( \frac{1}{2 - \cos(0.6843)} \right) \approx 0.6846\)

The iterations converge to approximately 0.68, confirming the root to 2 decimal places.