cos(x/4-п/6)=<-1/2 помогите пожалуйста</p>
cos(x/4-п/6)=<-1/2</span>
x/4-п/6=t
cos t <=-1/2</span>
t=2pi/3+2pik
t=4pi/3+2pik
Найдем х:
1)x/4-pi/6=2pi/3+2pik
x/4=5pi/6+2pik
x=10pi/3+8pik . k=z
2)x/4-pi/6=4pi/3+2pik
x/4=3pi/2+2pik
x=6pi+2pik=2pik . k=z
x=[10pi/3+8pik;2pik] . k=z