The given equation is:

$\cot^2 \theta - 3 \csc \theta + 3 = 0, \quad 0^\circ < \theta < 180^\circ$

To solve this equation, let's proceed by using trigonometric identities and algebraic manipulations.

1. Rewrite $\cot^2 \theta$ in terms of $\csc \theta$:
   Using the identity $\cot^2 \theta = \csc^2 \theta - 1$, substitute into the equation: $\csc^2 \theta - 1 - 3 \csc \theta + 3 = 0$ Simplify this: $\csc^2 \theta - 3 \csc \theta + 2 = 0$

2. Set up a substitution:
   Let $x = \csc \theta$. Then, the equation becomes: $x^2 - 3x + 2 = 0$

3. Solve the quadratic equation:
   Factorize the quadratic: $(x - 1)(x - 2) = 0$ So, $x = 1$ or $x = 2$.

4. Convert back to $\theta$:

   • For $x = 1$: $\csc \theta = 1 \Rightarrow \sin \theta = 1$, which gives $\theta = 90^\circ$.
   • For $x = 2$: $\csc \theta = 2 \Rightarrow \sin \theta = \frac{1}{2}$, which gives $\theta = 30^\circ$ and $\theta = 150^\circ$ within the range $0^\circ < \theta < 180^\circ$.

5. Final Solution: The values of $\theta$ that satisfy the equation are: $\theta = 30^\circ, \quad \theta = 90^\circ, \quad \theta = 150^\circ$

Would you like further details on any step or have any questions?

Here are five related questions that expand on this topic:

1. How can we solve similar trigonometric equations involving $\sec$ or $\tan$?
2. What are other common trigonometric identities useful for solving equations?
3. How would the solution change if the range for $\theta$ was different, such as $0 < \theta < 360$?
4. Could this equation have solutions in radians instead of degrees?
5. How can we verify each solution by substituting back into the original equation?

Tip: When solving trigonometric equations, always consider converting identities to simpler forms or use substitutions to make the equation easier to solve.