Cosx-sinx=2sin^2x/2 помогите))))

cosx - sinx = 2sin²(x/2) cos²(x/2) - sin²(x/2) - 2 sin(x/2)cos(x/2) - 2sin²(x/2) = 0 3sin²(x/2) + 2sin(x/2)cos(x/2) - cos²(x/2) = 0 делим на (cos²(x/2) ≠ 0) 3tg²(x/2) + 2tg(x/2) - 1 = 0 D = 4 + 4*3*1 = 16 1)  tg(x/2) = ( - 2 - 4)/6 tg(x/2) = - 1 x/2 = - π/4 + πn, n∈Z x₁ = - π/2 + 2πn, n∈Z 2)   tg(x/2) = ( - 2 + 4)/6 tg(x/2) = 1/3 x/2 = arctg(1/3) + πk, k∈Z) x₂ = 2*arctg(1/3) + 2πk, k∈Z)