Решить уравнение.

1-cos2x = 0.5√3tgx

1-cos2x=0,5√3tgx

1 - (1-2sin²x) = 0,√3tgx

2sin²x=0,5√3tgx

sin²x = (√3/4) tgx

sin²x - (√3/4) (sinx/cosx) = 0

six (sin-√3/4cosx) = 0

sinx=0 или sinx-√3/cosx=0

1. sinx=0

x₁=2πn, n∈Z

2. sinx-√3/4cosx=0

(4sinx*cosx-√3) / 4cosx=0

4sinx*cosx-√3=0, cos≠0

2 * (2sinxcosx) = √3

sin2x=√3/2

2x = (-1) ^n*arcsin (√3/2) + πn, n∈Z

2x = (-1) ^n * (π/3) + πn, n∈Z | : 2

x₂ = (-1) ^n * (π/6) + πn/2, n∈Z