1. **State the problem:** Solve the equation $$2\cot^2\theta + 3\csc^2\theta = 4\cot\theta + 3$$ for $$-180^\circ \leq \theta \leq 180^\circ$$. 2. **Recall identities:** We know that $$\csc^2\theta = 1 + \cot^2\theta$$. 3. **Substitute:** Replace $$\csc^2\theta$$ in the equation: $$2\cot^2\theta + 3(1 + \cot^2\theta) = 4\cot\theta + 3$$ 4. **Expand and simplify:** $$2\cot^2\theta + 3 + 3\cot^2\theta = 4\cot\theta + 3$$ $$5\cot^2\theta + 3 = 4\cot\theta + 3$$ 5. **Subtract 3 from both sides:** $$5\cot^2\theta + \cancel{3} - \cancel{3} = 4\cot\theta + \cancel{3} - \cancel{3}$$ $$5\cot^2\theta = 4\cot\theta$$ 6. **Rewrite as quadratic in $$\cot\theta$$:** $$5\cot^2\theta - 4\cot\theta = 0$$ 7. **Factor out $$\cot\theta$$:** $$\cot\theta(5\cot\theta - 4) = 0$$ 8. **Set each factor to zero:** - $$\cot\theta = 0$$ - $$5\cot\theta - 4 = 0 \Rightarrow \cot\theta = \frac{4}{5}$$ 9. **Solve $$\cot\theta = 0$$:** $$\cot\theta = \frac{\cos\theta}{\sin\theta} = 0 \Rightarrow \cos\theta = 0$$ Within $$-180^\circ \leq \theta \leq 180^\circ$$, $$\cos\theta=0$$ at $$\theta = \pm 90^\circ$$. 10. **Solve $$\cot\theta = \frac{4}{5}$$:** $$\cot\theta = \frac{4}{5} \Rightarrow \tan\theta = \frac{5}{4}$$ 11. **Find $$\theta$$:** $$\theta = \arctan\left(\frac{5}{4}\right) \approx 51.34^\circ$$ 12. **Find all solutions in the interval:** Since $$\tan\theta$$ has period $$180^\circ$$, solutions are: $$\theta \approx 51.34^\circ$$ and $$51.34^\circ - 180^\circ = -128.66^\circ$$. 13. **Final solutions:** $$\theta = -180^\circ, -128.66^\circ, 90^\circ, 51.34^\circ$$ (Note: $$-180^\circ$$ is not a solution here; only $$\pm 90^\circ$$ and the two $$\cot\theta=4/5$$ solutions.) **Answer:** $$\boxed{\theta = -128.66^\circ, -90^\circ, 90^\circ, 51.34^\circ}$$