Решить уравнение : ((sin2x+sqrt3cos2x)^2)-5=cos(pi/6-2x)

(2(1/2sin2x+√3/2cos2x))² -5 =sin(π/2-π/6+2x)
(2sin(2x+π/3))² -5 -sin(2x+π/3)=0
4sin²(2x+π/3)-sin(2x+π/3) -5=0
sin(2x+π/3)=f
4a²-a-5=0
D=1+80=81
a1=(1-9)/8=-1⇒sin(2x+π/3)=-1⇒2x+π/3=-π/2+2πn⇒2x=-π/3-π/2+2πn=-5π/6+2πn⇒
⇒x=-5π/12+πn
a2=(1+9)/8=5/4∉[-1;1]-нет решения