Question

решите неравенство:
sin X <корень из 2/2

Answer 1

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

Answer 2

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

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

Ответ: x ∈ (3π/4; 9π/4) + 2πn, n ∈ Z