1. Stating the problem: Solve the trigonometric equation $$\sin 2x \cos 2x - 2 \sin 2x = 0$$. 2. Factor the equation: $$\sin 2x (\cos 2x - 2) = 0$$ 3. Set each factor equal to zero and solve: 3.1. $$\sin 2x = 0$$ The solutions are: $$2x = n\pi \implies x = \frac{n\pi}{2} \quad \text{for integers } n$$ 3.2. $$\cos 2x - 2 = 0 \implies \cos 2x = 2$$ Since the cosine function ranges only from -1 to 1, $$\cos 2x = 2$$ has no real solutions. 4. Final solution: The solutions come only from $$\sin 2x = 0$$, $$\boxed{x = \frac{n\pi}{2} \quad \text{for all integers } n}$$