Verified answer

Ответ:

[tex]\dfrac{\pi }{2} + 4\pi n < x < \dfrac{7}{2}\pi + 4\pi n ~ , n \in \mathbb Z[/tex]

Объяснение:

[tex]\displaystyle f'(x) > 0 \\\\ (4 \sin \frac{x}{2} - x\sqrt{2})' > 0 \\\\ 4 \cdot \frac{1}{2} \cdot \cos \frac{x}{2 } - \sqrt{2} > 0 \\\\ \cos \frac{x}{2} > \frac{\sqrt{2} }{2} \\\\ \arccos \frac{\sqrt{2} }{2} + 2\pi n < \frac{x}{2} < 2\pi - \arccos \frac{\sqrt{2} }{2} +2 \pi n \\\\ \frac{\pi }{4} + 2 \pi n < \frac{x}{2} < 2\pi - \frac{\pi }{4} + 2\pi n \\\\\\ \dfrac{\pi }{2} + 4\pi n < x < \dfrac{7}{2}\pi + 4\pi n ~ , n \in \mathbb Z[/tex]