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