Помогите решить √3*sin x/2+1=cosx

√3*sin (x/2) + 1=cosx

cosx=cos (2 * (x/2)) = cos² (x/2) - sin² (x/2) = 1-2sin² (x/2)

√3*sin (x/2) + 1=1-2sin² (x/2)

2sin² (x/2) + √3*sin (x/2) = 0

sin (x/2) * (2sin (x/2) + √3) = 0

sin (x/2) = 0 или 2sin (x/2) + √3=0

1. sin (x/2) = 0 частный случай.

x/2=π/2+2πn, n∉Z |*2

x₁=π+4πn, n∈Z

2. 2sin (x/2) + √3=0

sin (x/2) = - √3/2

x/2 = (-1) ^n * arcsin (-√3/2) + πn, n∈Z

x/2 = (-1) ^ (n+1) * (π/3) + πn, n∈Z |*2

x₂ = (-1) ^ (n+1) * (2π/3) + 2πn, n∈Z