Question

cos(2x-2пи/3)+5sin(x-пи/3)+2=0

Answer 1

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.