Question

решите неравенство:
sin X <корень из 2/2

Answer 1

$-\frac{5\pi}{4} +2\pi k <x <\frac{\pi}{4} +2\pi k,$k∈Z

Answer 2

$\sin x < \dfrac{\sqrt{2}}{2} \\ \\ \left[ \begin{gathered} x < \dfrac{\pi}{4} + 2\pi n, n \in Z \\ x < \dfrac{3\pi}{4} + 2\pi n, n \in Z\\ \end{gathered} \right.$

$x \in (\dfrac{3\pi}{4} ; \dfrac{9\pi}{4}) + 2\pi n, n \in Z$

Ответ: x ∈ (3π/4; 9π/4) + 2πn, n ∈ Z