1. **State the problem:** Solve the trigonometric equation $$\frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0.$$\n\n2. **Recall formulas and identities:**\n- The cotangent function is defined as $$\cot \theta = \frac{\cos \theta}{\sin \theta}.$$\n- Factor powers of cosine and sine where possible.\n\n3. **Rewrite the numerator and denominator:**\n- Numerator: $$\cos \theta - 2 \cos^3 \theta = \cos \theta (1 - 2 \cos^2 \theta).$$\n- Denominator: $$\sin \theta - 2 \sin^3 \theta = \sin \theta (1 - 2 \sin^2 \theta).$$\n\n4. **Rewrite the fraction:**\n$$\frac{\cos \theta (1 - 2 \cos^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)} + \frac{\cos \theta}{\sin \theta} = 0.$$\n\n5. **Combine the terms over a common denominator:**\n$$\frac{\cos \theta (1 - 2 \cos^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)} + \frac{\cos \theta (1 - 2 \sin^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)} = 0.$$\n\n6. **Add the numerators:**\n$$\frac{\cos \theta \left[(1 - 2 \cos^2 \theta) + (1 - 2 \sin^2 \theta)\right]}{\sin \theta (1 - 2 \sin^2 \theta)} = 0.$$\n\n7. **Simplify inside the bracket:**\n$$ (1 - 2 \cos^2 \theta) + (1 - 2 \sin^2 \theta) = 2 - 2(\cos^2 \theta + \sin^2 \theta) = 2 - 2(1) = 0.$$\n\n8. **Therefore, numerator is:**\n$$\cos \theta \times 0 = 0.$$\n\n9. **The fraction equals zero when numerator is zero and denominator is not zero:**\n- Numerator zero: $$\cos \theta = 0.$$\n- Denominator not zero: $$\sin \theta (1 - 2 \sin^2 \theta) \neq 0.$$\n\n10. **Solve for $$\theta$$:**\n- $$\cos \theta = 0 \implies \theta = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}.$$\n- Check denominator at these points:\n - $$\sin \theta \neq 0$$ (true at $$\theta = \frac{\pi}{2} + k\pi$$ since sine alternates between 1 and -1).\n - $$1 - 2 \sin^2 \theta \neq 0$$\n\n11. **Check denominator term $$1 - 2 \sin^2 \theta = 0$$:**\n$$1 - 2 \sin^2 \theta = 0 \implies \sin^2 \theta = \frac{1}{2} \implies \sin \theta = \pm \frac{\sqrt{2}}{2}.$$\n\n12. **At $$\theta = \frac{\pi}{2} + k\pi$$, $$\sin \theta = \pm 1$$, so denominator is not zero.**\n\n**Final answer:**\n$$\boxed{\theta = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}}.$$