7. Решите уравнение: $$\sin 2x + 2 \cos(\frac{\pi}{2} - x) = \sqrt{3} \cos x + \sqrt{3}$$

$$2 \sin x \cos x + 2 \sin x = \sqrt{3} \cos x + \sqrt{3} \implies 2 \sin x (\cos x + 1) = \sqrt{3} (\cos x + 1)$$. $$(2 \sin x - \sqrt{3})(\cos x + 1) = 0$$. $$\sin x = \frac{\sqrt{3}}{2}$$ or $$\cos x = -1$$. $$x = \frac{\pi}{3} + 2\pi k$$, $$x = \frac{2\pi}{3} + 2\pi k$$, $$x = \pi + 2\pi k$$, $$k \in \mathbb{Z}$$.