Cos(2x-2пи/3)+5sin(x-пи/3)+2=0

cos(2x-2π/3) +5sin(x-π/3)+2=0 cos2(x-π/3) +5sin(x-π/3)+2=0 cos²(x-π/3) - sin²(x-π/3) +5sin(x-π/3)+2=0 1-sin²(x-π/3) -sin²(x-π/3) +5sin(x-π/3) +2 =0 1 -2sin²(x-π/3) +5sin(x-π/3) +2=0 2sin²(x-π/3) - 5sin(x-π/3) -3 =0 sin(x-π/3)=y 2y² - 5y -3=0 D=25 +24=49 y₁=5 -7 = -1/2        4 y₂ =5+7 =3         4 При у= -1/2 sin(x -π/3) = -1/2 x-π/3 =(-1)^(n+1) * (π/6) +πn, n∈Z x=(-1)^(n+1) * (π/6) + π/3 +πn, n∈Z При у=3 sin(x-π/3)=3 Так как 3∉[-1; 1], то  уравнение не имеет корней. Ответ: (-1)^(n+1) * (π/6) +π/3 +πn, n∈Z.