А) sinx + 2sinx (2x + π/6) = √3sin2x + 2

б) корни на отрезке [-7π/2; - 2π]

Sinx+2sin (2 х+π/6) = √3sin2x+2

sinx+2 (sin2x•cosπ/6+cos2x•sinπ/6) =

√3sin2x+2

sinx+√3sin2x+cos2x=√3sin2x+2

sinx+cos2x-2=0

sinx+cos²x-sin²x-2=0

sinx+1-2sin²x-2=0

2sin²x-sinx-1=0

sinx=t

2t²-t+1=0

D=1+8=9

t = (1±3) / 4

t1=1; t2=-1/2

sinx=1; x=π/2+2πk

sinx=-1/2; x = (-1) ⁿ (-5π/6) + πk; k€Z