2 sin во 2ст x +3 cos x = 0 помогите)!

2sin²x+3cosx=02*(1-cos²x)+3cosx=0-2cos²x+3cosx+2=0cosx=t, t∈[-1;1]2t²-3t-2=0t₁=2, 2∉[-1;1]t₂=-1/2t=-1/2cosx=-1/2x=+-(arccos(-1/2))+2πn, n∈Zx=+-(π-arccos(1/2))+2πn, n∈Zx=+-(2π/3)+2πn, n∈Z