4. Решить неравенство: 1) cosx ≤ -√3/2; 2) sin(2-3x) > √2/2.

Решение:

1. \( \cos x \le -\frac{\sqrt{3}}{2} \)
   \( x \in \left[ \frac{5\pi}{6} + 2\pi n; \frac{7\pi}{6} + 2\pi n \right], n \in \mathbb{Z} \)
2. \( \sin (2-3x) > \frac{\sqrt{2}}{2} \)
   \( \frac{\pi}{4} + 2\pi k < 2-3x < \frac{3\pi}{4} + 2\pi k, k \in \mathbb{Z} \)
   \( \frac{\pi}{4} - 2 - 2\pi k < -3x < \frac{3\pi}{4} - 2 - 2\pi k \)
   \( 2 + 2\pi k - \frac{\pi}{4} > 3x > 2 + 2\pi k - \frac{3\pi}{4} \)
   \( \frac{2}{3} + \frac{2\pi k}{3} - \frac{\pi}{12} > x > \frac{2}{3} + \frac{2\pi k}{3} - \frac{\pi}{4} \)
   \( x \in \left( \frac{2}{3} - \frac{\pi}{4} + \frac{2\pi k}{3}; \frac{2}{3} - \frac{\pi}{12} + \frac{2\pi k}{3} \right), k \in \mathbb{Z} \)

Ответ: 1) \( x \in \left[ \frac{5\pi}{6} + 2\pi n; \frac{7\pi}{6} + 2\pi n \right], n \in \mathbb{Z} \); 2) \( x \in \left( \frac{2}{3} - \frac{\pi}{4} + \frac{2\pi k}{3}; \frac{2}{3} - \frac{\pi}{12} + \frac{2\pi k}{3} \right), k \in \mathbb{Z} \).