(a) Sketch the graph of \(y = 4 - x^2\) and \(y = \sec \frac{1}{2}x\). The intersection of these graphs in the interval \(0 \leq x < \pi\) shows the root. The graph of \(y = 4 - x^2\) is a downward-opening parabola with vertex at (0, 4) and intersects the x-axis at \(x = 2\) and \(x = -2\). The graph of \(y = \sec \frac{1}{2}x\) has a vertical asymptote at \(x = \pi\) and intersects the y-axis at (0, 1). The graphs intersect exactly once in the interval \(0 \leq x < \pi\).

(b) Calculate \(4 - x^2\) and \(\sec \frac{1}{2}x\) at \(x = 1\) and \(x = 2\):

At \(x = 1\), \(4 - 1^2 = 3\) and \(\sec \frac{1}{2}(1) = \sec 0.5 \approx 1.139\).

At \(x = 2\), \(4 - 2^2 = 0\) and \(\sec \frac{1}{2}(2) = \sec 1 \approx 1.850\).

Since \(3 > 1.139\) and \(0 < 1.850\), the root lies between 1 and 2.

(c) Use the iterative formula \(x_{n+1} = \sqrt{4 - \sec \frac{1}{2}x_n}\):

Start with \(x_0 = 1.5\).

\(x_1 = \sqrt{4 - \sec \frac{1}{2}(1.5)} \approx 1.6161\)

\(x_2 = \sqrt{4 - \sec \frac{1}{2}(1.6161)} \approx 1.6025\)

\(x_3 = \sqrt{4 - \sec \frac{1}{2}(1.6025)} \approx 1.6003\)

\(x_4 = \sqrt{4 - \sec \frac{1}{2}(1.6003)} \approx 1.6000\)

The root correct to 2 decimal places is 1.60.