1. **Stating the problem:** Solve the equation $$6 \cot x \left(1 + \cot^2 x\right) = \frac{3\sqrt{3}}{2} \sin 2x$$ for $$0 \leq x \leq 2\pi$$. 2. **Recall identities:** - $$1 + \cot^2 x = \csc^2 x$$. - $$\sin 2x = 2 \sin x \cos x$$. - Also, $$\cot x = \frac{\cos x}{\sin x}$$. 3. **Rewrite the left side:** $$6 \cot x (1 + \cot^2 x) = 6 \cot x \csc^2 x = 6 \frac{\cos x}{\sin x} \frac{1}{\sin^2 x} = 6 \frac{\cos x}{\sin^3 x}$$. 4. **Rewrite the right side:** $$\frac{3\sqrt{3}}{2} \sin 2x = \frac{3\sqrt{3}}{2} \cdot 2 \sin x \cos x = 3\sqrt{3} \sin x \cos x$$. 5. **Set the equation:** $$6 \frac{\cos x}{\sin^3 x} = 3\sqrt{3} \sin x \cos x$$. 6. **Divide both sides by $$\cos x$$ (assuming $$\cos x \neq 0$$):** $$6 \frac{1}{\sin^3 x} = 3\sqrt{3} \sin x$$. 7. **Multiply both sides by $$\sin^3 x$$:** $$6 = 3\sqrt{3} \sin^4 x$$. 8. **Divide both sides by $$3\sqrt{3}$$:** $$\frac{6}{3\sqrt{3}} = \sin^4 x \implies \frac{2}{\sqrt{3}} = \sin^4 x$$. 9. **Simplify $$\frac{2}{\sqrt{3}}$$:** $$\frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.1547$$. 10. **Analyze:** Since $$\sin^4 x \leq 1$$ for all real $$x$$, there is no real solution where $$\sin^4 x = 1.1547$$. 11. **Check the case $$\cos x = 0$$:** - If $$\cos x = 0$$, then $$x = \frac{\pi}{2}, \frac{3\pi}{2}$$. - Substitute into original equation: - Left side: $$6 \cot x (1 + \cot^2 x)$$ is undefined because $$\cot x = \frac{\cos x}{\sin x} = 0$$ in denominator. - So no solution here. **Final conclusion:** No real solutions exist for $$x$$ in $$[0, 2\pi]$$ satisfying the given equation.