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

sin2x=(cos²x/2-sin²x/2)(cos²x/2+sin²x/2)sin2x=cosx2sinxcosx-cosx=0cosx(2sinx-1)=0cosx=0⇒x=π/2+πnsinx=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