Question

Решите неравенство 2sinx≤1
варианты:
[-4π/3+2πn; пи/3+2πn] , n∈Z
[π/3+2πn; 2π/3+2πn], n∈Z
[-7π/6+2πn; π/6+2πn], n∈Z
[π/6+2πn; 5π/6+2πn], n∈Z

Answer 1

2sinx≤1
sinx≤0,5
-π-π/6≤x≤π/6
-7π/6+2πk≤x≤π/6+2πk,k∈z
[-7π/6+2πk;π/6+2πk],k∈z