Question

решите уравнение, желательно с рисунком плиз sin3x=корень из 3/2 на промежутке от -3П/2 до П

Answer 1

sin3x=√3/2
3x=π/3 +2πn,n∈Z  U 3x=2π/3+2πk,k∈Z
x=π/9+2πn/3,n∈Z U x=2π/9+2πk,k∈Z
1)-3π/2≤π/9+2πn/3≤π 
-29≤12n≤16
-2 5/12≤n≤1 1/3
n=-2⇒x=π/9-4π/3=-11π/9
n=-1⇒x=π/9-2π/3=-5π/9
n=0⇒x=π/9
n=1⇒x=π/9+2π/3=7π/9
2)  -3π/2≤2π/9+2πk/3≤π 
-31≤12k≤14
-2 7/12≤k≤1 1/6
k=-2⇒x=2π/9-4π/3=-10π/9
k=-1⇒x=2π/9-2π/3=-4π/9
k=0⇒x=2π/9
k=1⇒x=2π/9+2π/3=8π/9