2(sin2xcosπ/6+sinπ/6cos2x)+1=√3sin2x+cosx
2sin2x·√3/2+2cos2x·1/2+1=√3sin2x+cosx
√3sin2x+cos2x+1=√3sin2x+cosx
cos2x+1-cosx=0
2cos²x-1+1-cosx=0
2cos²x-cosx=0
cosx(2cosx-1)=0
cosx=0 x=π/2+πn n принадлежит Z
2cosx-1=0
cosx=1/2 x=π/3+2πn x=-π/3+2πn n принадлежит Z