Question

Неравенства
2sin*3x>0
и второе 2sin*3x<0

Answer 1

sinx>0⇒x∈(2πn;π+2πn)
sinx<0⇒x∈(π+2πn;2π+2πn)
1)
2sin3x>0
sin3x>0
3x∈(2πn;π+2πn)
x∈(2πn/3;π/3+2πn/3)
2)
2sin3x<0
sin3x<0
3x∈(π+2πn;2π+2πn)
x∈(π/3+2πn/3;2π/3+2πn/3)