(a) For \(0\le x<\pi/4\), the secant graph is positive and the line negative, so they do not meet. For \(\pi/4<x\le\pi/2\), \(\sec2x\) increases from \(-\infty\) to \(-1\) while the line decreases; hence they cross exactly once. The asymptote \(x=\pi/4\) is excluded:

(b) Let \(F(x)=\sec2x+2x+1/2\). On \([0.8,1.2]\) it is continuous, with \[F(0.8)=-32.1<0,\qquad F(1.2)=1.54>0.\] Thus the root lies between them.

(c) By continuity at a real convergent limit \(L\), \[L=\frac12\cos^{-1}\left(\frac{-2}{4L+1}\right)\Rightarrow\cos2L=\frac{-2}{4L+1}\Rightarrow\sec2L=-2L-\frac12.\] The inverse cosine puts the limit in \([0,\pi/2]\). The denominator is positive there and the cosine negative, so the limit lies to the right of \(\pi/4\); by (a) it is the unique root.

(d) Start at \(x_0=1\) and retain full precision:

The root is \(x=0.99248\ldots=0.992\) to 3 decimal places.