Решение
2sin^2(x)=3cos(x)
2*(1 - cos²x) - 3cosx = 0
2 - 2cos²x - 3cosx = 0
2cos²x + 3cosx - 2 = 0
cosx = t, I t I ≤ 1
2t² + 3t - 2 = 0
D = 9 + 4*2*2 = 25
t₁ = (- 3 - 5)/4 = - 2, не удовлетворяет условию I t I ≤ 1
t₂ = (- 3 + 5)/4 = 1/2
cosx = 1/2
x = (+ -) arccos(1/2) + 2πk, k ∈ z
x = (+ -) π/3 + 2πk, k ∈ z