Решите уравнение sin2x=cos^4*x/2-sin^4*x/2

Sin2x = (cos²x/2-sin²x/2) (cos²x/2+sin²x/2)

sin2x=cosx

2sinxcosx-cosx=0

cosx (2sinx-1) = 0

cosx=0⇒x=π/2+πn

sinx=1/2⇒x = (-1) ^n * π/6+πn

In2x = cos⁴ (x/2) - sin⁴ (x/2) sin2x = (cos² (x/2) + sin² (x/2)) (cos² (x/2) - sin² (x/2)) sin2x = cos² (x/2) - sin² (x/2) sin2x = cosxsin2x = sin (π/2-x) 2x = π/2-x+2πn

n∈Z ∨ 2x = π - π/2 + x + 2πn

n∈Zx₁ = ⅙ (4πn + π)

n∈Z ∨ x₂ = ½ (4πn + π)

n∈Z