Решите уравнение sin2x=sin4x

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sin2x-sin4x=0sin2x - 2 sin2x*cos2x=0sin2x (1-2 cos2x) =0sin 2x =0 ⇒2x=πn, n∈Z ⇒ x₁ = πn\ 2, n∈Zили1-2 cos2x=0⇒cos2x=1\2 ⇒2x=⁺- arccos 1\2+πn, n∈Z; 2x=⁺-π\3+2πn, n∈Zx₂= ⁺-π\6+πn, n∈Z