To solve for $\theta$ in the given interval $[0^\circ, 360^\circ]$ based on the equation:

$\csc \theta = \frac{2 \sqrt{3}}{3}$

We first recognize that $\csc \theta = \frac{1}{\sin \theta}$, so:

$\frac{1}{\sin \theta} = \frac{2 \sqrt{3}}{3}$

Now, solve for $\sin \theta$:

$\sin \theta = \frac{3}{2 \sqrt{3}}$

Simplifying $\sin \theta$:

$\sin \theta = \frac{\sqrt{3}}{2}$

Now, we need to find all values of $\theta$ such that $\sin \theta = \frac{\sqrt{3}}{2}$ in the interval $[0^\circ, 360^\circ]$.

The general angles where $\sin \theta = \frac{\sqrt{3}}{2}$ are:

$\theta = 60^\circ \quad \text{(in the first quadrant)}$ and $\theta = 180^\circ - 60^\circ = 120^\circ \quad \text{(in the second quadrant)}.$

Thus, the solutions for $\theta$ are:

$\theta = 60^\circ \quad \text{and} \quad \theta = 120^\circ.$

Let me know if you want any further details or have questions! Here are five related questions:

1. What is the value of $\sin(60^\circ)$ and why does it equal $\frac{\sqrt{3}}{2}$?
2. How do the reciprocal trigonometric functions like $\csc$ relate to $\sin$?
3. Can we find other values of $\theta$ if the interval changes to $[0^\circ, 720^\circ]$?
4. What happens to the angle solutions if we work in radians instead of degrees?
5. How can the Pythagorean identity help verify the solutions for $\sin \theta = \frac{\sqrt{3}}{2}$?

Tip: Memorizing the values of trigonometric functions for common angles like $30^\circ$, $45^\circ$, and $60^\circ$ is useful for solving many trigonometric equations efficiently.