2πх=(π/6)+πk, k∈Z
Делим на 2π
х=(1/12)+(k/2)
0 < (1/12)+(k/2)<3<br>0 < 1+6k < 36
-1< 6k < 35
k=1; 2; 3; 4; 5
x=(1/12)+(1/2)=7/12
x=(1/12)+(2/2)=13/12
x=(1/12)+(3/2)=19/12
x=(1/12)+(4/2)=25/12
x=(1/12)+(5/2)=31/12
О т в е т. 7/12; 13/12; 19/12; 25/12; 31/12.