Question

а) -√3-2sin3x<0

б) ctg ($frac{ pi }{2}$+x)>√3

Answer 1

sin3x>-√3/2
-π/3+2πn<3x<4π/3+2πn
-π/9+2πn/3<x<4π/9+2πn/3
x∈(-π/9+2πn/3;4π/9+2πn/3)

πn<π/2+x<π/6+πn
-π/2+πn<x<-π/2+π/6+πn
-π/2+πn<x<-π/3+πn
x∈(-π/2+πn;-π/3+πn)