10) sinxcosx=√2 4

Решим уравнение $$sinxcosx = \frac{\sqrt{2}}{4}$$

$$2sinxcosx = \frac{\sqrt{2}}{2}$$

$$sin2x = \frac{\sqrt{2}}{2}$$

$$2x = arcsin(\frac{\sqrt{2}}{2}) + 2\pi n, n \in Z$$ или $$2x = \pi - arcsin(\frac{\sqrt{2}}{2}) + 2\pi n, n \in Z$$

$$2x = \frac{\pi}{4} + 2\pi n, n \in Z$$ или $$2x = \frac{3\pi}{4} + 2\pi n, n \in Z$$

$$x = \frac{\pi}{8} + \pi n, n \in Z$$ или $$x = \frac{3\pi}{8} + \pi n, n \in Z$$

Ответ: $$x = \frac{\pi}{8} + \pi n, n \in Z$$ или $$x = \frac{3\pi}{8} + \pi n, n \in Z$$