Делим на 2sqrt(3)
(1/2)sin2x-(sqrt(3)/2) cos2x ≥ 1 ⇒
(1/2)=sin(π/6)
(sqrt(3))/2=cos(π/6)
sin(π/6)*sin2x-cos(π/6)*cos2x ≥ 1
По формуле cos α cos β -sin α sin β =cos( α + β )
-cos(2x+(π/6)) ≥ 1 ⇒ cos(2x+(π/6)) ≤ -1 ⇒ cos(2x+(π/6)=-1
2x+(π/6)=π+2πn, n ∈ Z
2x=(5π/6)+2πn, n ∈ Z
[b]x=(5π/12)+πn, n ∈ Z[/b]