1. **State the problem:** Solve the equation $$\cot 2x = - \sin x \cdot \cos x$$. 2. **Recall the formulas:** - $$\cot 2x = \frac{\cos 2x}{\sin 2x}$$ - Double angle formulas: $$\sin 2x = 2 \sin x \cos x$$ $$\cos 2x = \cos^2 x - \sin^2 x$$ 3. **Rewrite the equation using these formulas:** $$\frac{\cos 2x}{\sin 2x} = - \sin x \cos x$$ 4. **Substitute double angle expressions:** $$\frac{\cos^2 x - \sin^2 x}{2 \sin x \cos x} = - \sin x \cos x$$ 5. **Multiply both sides by $$2 \sin x \cos x$$ to clear the denominator:** $$\cancel{2 \sin x \cos x} \cdot \frac{\cos^2 x - \sin^2 x}{\cancel{2 \sin x \cos x}} = - \sin x \cos x \cdot 2 \sin x \cos x$$ Simplifies to: $$\cos^2 x - \sin^2 x = - 2 \sin^2 x \cos^2 x$$ 6. **Rewrite the left side as $$\cos 2x$$:** $$\cos 2x = - 2 \sin^2 x \cos^2 x$$ 7. **Express $$\sin^2 x \cos^2 x$$ in terms of double angle:** Recall: $$\sin 2x = 2 \sin x \cos x \implies \sin^2 2x = 4 \sin^2 x \cos^2 x$$ So: $$\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}$$ 8. **Substitute back:** $$\cos 2x = - 2 \cdot \frac{\sin^2 2x}{4} = - \frac{\sin^2 2x}{2}$$ 9. **Multiply both sides by 2:** $$2 \cos 2x = - \sin^2 2x$$ 10. **Rewrite as:** $$2 \cos 2x + \sin^2 2x = 0$$ 11. **Use identity $$\sin^2 \theta = 1 - \cos^2 \theta$$:** $$2 \cos 2x + 1 - \cos^2 2x = 0$$ 12. **Rearrange:** $$- \cos^2 2x + 2 \cos 2x + 1 = 0$$ Multiply both sides by -1: $$\cos^2 2x - 2 \cos 2x - 1 = 0$$ 13. **Let $$y = \cos 2x$$, solve quadratic:** $$y^2 - 2y - 1 = 0$$ 14. **Use quadratic formula:** $$y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$ 15. **Check valid values:** Since $$\cos 2x$$ must be between -1 and 1, - $$1 + \sqrt{2} > 1$$ invalid - $$1 - \sqrt{2} < -1$$ invalid No valid solutions from this quadratic. 16. **Check for domain restrictions:** Recall original equation denominator $$\sin 2x \neq 0$$. 17. **Re-examine step 5:** Multiplying by $$2 \sin x \cos x$$ assumes $$\sin x \neq 0$$ and $$\cos x \neq 0$$. 18. **Check cases where $$\sin x = 0$$ or $$\cos x = 0$$:** - If $$\sin x = 0$$, then $$\cot 2x = -0 = 0$$, but $$\cot 2x = \frac{\cos 2x}{\sin 2x}$$ undefined if $$\sin 2x = 0$$. - If $$\cos x = 0$$, similar check. 19. **Check $$\cot 2x = - \sin x \cos x$$ directly for $$x = k \frac{\pi}{2}$$:** - For $$x = \frac{\pi}{2}$$, $$\sin x = 1$$, $$\cos x = 0$$, RHS = 0 - $$\cot 2x = \cot \pi = \infty$$ undefined No solution here. 20. **Conclusion:** No real solutions satisfy the equation. **Final answer:** $$\boxed{\text{No real solutions}}$$