Question

cosx-sinx=2sin^2x/2 помогите))))

Answer 1

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)