Cos(2x - 2π/3) = Cos(2(x - π/3)) = 1 - 2Sin²(x - π/3)
теперь наше уравнение:
Sin(x-π/3)=1 - 2Sin²(x - π/3)
2Sin²(x - π/3) +Sin(x-π/3) -1 = 0
Sin(x-π/3)=t
2t² +t -1 = 0
D = 9
t₁=(-1-3)/4 = -1; t₂=(-1 +3)/4 = 1/2
a) t₁ = 1
Sin(x-π/3)=1
x-π/3 = π/2 + 2πk, k ∈ Z
x = π/3 + π/2 + 2πk, k ∈ Z
x = 5π/6 + 2πk, k ∈ Z
б) t₂ = 1/2
Sin(x-π/3)= 1/2
x-π/3 = (-1)^n * π/6 + nπ, n ∈ Z
x = π/3 + (-1)^n * π/6 + nπ, n ∈ Z