Решите уравнение

2cos^2 x - 3sin x=0

2cos²x-3sinx=0

2 * (1-sin²x) - 3sinx=0

2sin²x+3sinx-2=0

sinx=t, t∈[-1; 1]

2t²+3t-2=0

D=3²-4*2 * (-2) = 25

t₁ = (-3-5) / 4, t₁=-2, - 2∉[-1; 1]

t₂ = (-3+5) / 4, t₂=1/2

sinx=1/2

x = (-1) ^n*arcsin (1/2) + πn, n∈Z

x = (-1) ^n * (π/6) + πn, n∈Z