Inverse Cosecant Function

Approach: We need an angle $\theta \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$ with $\csc \theta = \frac{2\sqrt{3}}{3}$, that is, with $\sin \theta = \frac{\sqrt{3}}{2}$.

Steps:

1. Among the special angles, $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$, so $\csc \frac{\pi}{3} = \frac{2\sqrt{3}}{3}$;
2. $\frac{\pi}{3} \approx 1.047$, which lies in $(0, \frac{\pi}{2}]$;
3. By uniqueness on the principal interval, $\operatorname{arccsc} \frac{2\sqrt{3}}{3} = \frac{\pi}{3}$.

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

Approach:

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

Steps:

1. The cosecant function $y = \csc x = \frac{1}{\sin x}$ 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 $y = \operatorname{arccsc} x$ is $(-\infty, -1] \cup [1, +\infty)$.
2. On $[-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$ the cosecant 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 cosecant function, so the range of $y = \operatorname{arccsc} x$ is $[-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$.

Answer:

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