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

2sin²x+3cosx=0

2 * (1-cos²x) + 3cosx=0

-2cos²x+3cosx+2=0

cosx=t, t∈[-1; 1]

2t²-3t-2=0

t₁=2, 2∉[-1; 1]

t₂=-1/2

t=-1/2

cosx=-1/2

x = + - (arccos (-1/2)) + 2πn, n∈Z

x = + - (π-arccos (1/2)) + 2πn, n∈Z

x = + - (2π/3) + 2πn, n∈Z