(i) Sketch the graphs of \(y = \csc x\) and \(y = \frac{1}{2}x + 1\) over the interval \(0 < x < \frac{1}{2}\pi\). The intersection of these graphs indicates a root in this interval.

(ii) Evaluate \(\csc x - \frac{1}{2}x - 1\) at \(x = 0.5\) and \(x = 1\). If the signs are different, a root exists between these values.

(iii) Rearrange \(\csc x = \frac{1}{2}x + 1\) to \(x = \sin^{-1} \left( \frac{2}{x+2} \right)\) to show equivalence.

(iv) Use the iterative formula \(x_{n+1} = \sin^{-1} \left( \frac{2}{x_n+2} \right)\) starting with \(x_1 = 0.75\). Calculate successive iterations until the value stabilizes to 2 decimal places: \(x_2 = 0.8041\), \(x_3 = 0.8002\), \(x_4 = 0.8000\). The root is approximately 0.80.