(i) Sketch the graphs of \(y = \sec x\) and \(y = 3 - \frac{1}{2}x^2\) over the interval \(0 < x < \frac{1}{2}\pi\). The intersection of these graphs indicates a root. The sketch should show that the graphs intersect, confirming a root exists in the interval.

(ii) Calculate \(\sec 1 - (3 - \frac{1}{2} \times 1^2)\) and \(\sec 1.4 - (3 - \frac{1}{2} \times 1.4^2)\). If the signs of these values are different, a root exists between 1 and 1.4.

(iii) Rearrange \(\sec x = 3 - \frac{1}{2}x^2\) to \(x = \cos^{-1}\left( \frac{2}{6 - x^2} \right)\) to show equivalence.

(iv) Use the iterative formula \(x_{n+1} = \cos^{-1}\left( \frac{2}{6 - x_n^2} \right)\) starting with an initial guess. Iterate until the value stabilizes to 4 decimal places. The iterations yield \(x \approx 1.13\).