Как решить 3sinx - 2cos^x = - 3?

3sinx - 2cos²x = - 3

3sinx - 2 (1-sin²x) = - 3

3sinx - 2 + 2sin²x = - 3

2sin²x + 3sinx + 1 = 0

sinx = t

2t² + 3t + 1 = 0

D = 9-8 = 1

√D = 1

x₁ = (-3-1) / 4 = - 1

x₂ = (-3+1) / 4 = - 1/2

sinx = - 1

sinx = - 1/2

x = - π/2 + 2πn

x = - π/6 + 2πn

x = - 5π/6 + 2πn

n∈Z