1. **State the problem:** Simplify and solve the expression $$\frac{\cos(3A) + \sin(A)}{\cos(3A) - \cos(A)} + \tan A = -2 \cot(2A)$$ for angle $A$. 2. **Recall formulas and identities:** - Triple angle formula for cosine: $$\cos(3A) = 4\cos^3 A - 3\cos A$$ - Double angle formula for cotangent: $$\cot(2A) = \frac{\cos(2A)}{\sin(2A)}$$ - Tangent and cotangent definitions: $$\tan A = \frac{\sin A}{\cos A}, \quad \cot A = \frac{\cos A}{\sin A}$$ - Double angle formulas: $$\cos(2A) = \cos^2 A - \sin^2 A, \quad \sin(2A) = 2\sin A \cos A$$ 3. **Rewrite numerator and denominator of the fraction:** $$\cos(3A) + \sin A = (4\cos^3 A - 3\cos A) + \sin A$$ $$\cos(3A) - \cos A = (4\cos^3 A - 3\cos A) - \cos A = 4\cos^3 A - 4\cos A = 4\cos A (\cos^2 A - 1)$$ 4. **Simplify denominator using Pythagorean identity:** $$\cos^2 A - 1 = -\sin^2 A$$ So denominator becomes: $$4\cos A (-\sin^2 A) = -4 \cos A \sin^2 A$$ 5. **Rewrite the fraction:** $$\frac{4\cos^3 A - 3\cos A + \sin A}{-4 \cos A \sin^2 A}$$ 6. **Rewrite the equation:** $$\frac{4\cos^3 A - 3\cos A + \sin A}{-4 \cos A \sin^2 A} + \frac{\sin A}{\cos A} = -2 \frac{\cos(2A)}{\sin(2A)}$$ 7. **Rewrite right side using double angle formulas:** $$-2 \frac{\cos(2A)}{\sin(2A)} = -2 \frac{\cos^2 A - \sin^2 A}{2 \sin A \cos A} = - \frac{\cos^2 A - \sin^2 A}{\sin A \cos A}$$ 8. **Multiply both sides by common denominator $-4 \cos A \sin^2 A$ to clear fractions:** Left side numerator times $-4 \cos A \sin^2 A$ is: $$4\cos^3 A - 3\cos A + \sin A$$ Left side second term multiplied: $$\frac{\sin A}{\cos A} \times (-4 \cos A \sin^2 A) = -4 \sin^3 A$$ Right side multiplied: $$- \frac{\cos^2 A - \sin^2 A}{\sin A \cos A} \times (-4 \cos A \sin^2 A) = 4 \sin A (\cos^2 A - \sin^2 A)$$ 9. **Write the equation after clearing denominators:** $$4\cos^3 A - 3\cos A + \sin A - 4 \sin^3 A = 4 \sin A (\cos^2 A - \sin^2 A)$$ 10. **Expand right side:** $$4 \sin A \cos^2 A - 4 \sin^3 A$$ 11. **Bring all terms to one side:** $$4\cos^3 A - 3\cos A + \sin A - 4 \sin^3 A - 4 \sin A \cos^2 A + 4 \sin^3 A = 0$$ 12. **Simplify terms:** $$-4 \sin^3 A + 4 \sin^3 A = 0$$ So equation reduces to: $$4\cos^3 A - 3\cos A + \sin A - 4 \sin A \cos^2 A = 0$$ 13. **Group terms:** $$4\cos^3 A - 3\cos A - 4 \sin A \cos^2 A + \sin A = 0$$ 14. **Factor where possible:** Group cosine terms: $$\cos A (4 \cos^2 A - 3) - \sin A (4 \cos^2 A - 1) = 0$$ 15. **Rewrite as:** $$(4 \cos^2 A - 3) \cos A = (4 \cos^2 A - 1) \sin A$$ 16. **Divide both sides by $\cos A$ (assuming $\cos A \neq 0$):** $$\cancel{\cos A} (4 \cos^2 A - 3) = (4 \cos^2 A - 1) \sin A / \cancel{\cos A}$$ So: $$4 \cos^2 A - 3 = (4 \cos^2 A - 1) \tan A$$ 17. **Solve for $\tan A$:** $$\tan A = \frac{4 \cos^2 A - 3}{4 \cos^2 A - 1}$$ 18. **Use identity $\cos^2 A = \frac{1}{1 + \tan^2 A}$ to express in terms of $\tan A$ or solve numerically depending on context.** **Final simplified relation:** $$\boxed{\tan A = \frac{4 \cos^2 A - 3}{4 \cos^2 A - 1}}$$ This expresses $\tan A$ in terms of $\cos^2 A$ consistent with the original equation.