Arcsecant Function

Approach: We need an angle $\theta \in [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$ with $\sec \theta = 2$, that is, with $\cos \theta = \frac{1}{2}$.

Steps:

1. Among the special angles, $\cos \frac{\pi}{3} = \frac{1}{2}$, so $\sec \frac{\pi}{3} = 2$;
2. $\frac{\pi}{3} \approx 1.047$, which lies in $[0, \frac{\pi}{2})$;
3. By uniqueness on the principal interval, $\operatorname{arcsec} 2 = \frac{\pi}{3}$.

Answer: $\operatorname{arcsec} 2 = \frac{\pi}{3}$.

Approach:

The arcsecant is the inverse of the secant function restricted to its principal interval $[0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$: the domain comes from the range of $\sec x$, and the range is the principal interval itself.

Steps:

1. The secant function $y = \sec x = \frac{1}{\cos x}$ has range $(-\infty, -1] \cup [1, +\infty)$, and the domain of an inverse function is exactly the range of the original function, so the domain of $y = \operatorname{arcsec} x$ is $(-\infty, -1] \cup [1, +\infty)$.
2. On $[0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$ the secant function is monotone on each of the two connected branches and the two ranges do not overlap, so this interval is a valid principal interval.
3. The inverse maps the angles of the principal interval back onto the range of the secant function, so the range of $y = \operatorname{arcsec} x$ is $[0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$.

Answer:

Domain: $(-\infty, -1] \cup [1, +\infty)$; Range: $[0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$.