Question

Решите уравнение sin2x=sin4x

Answer 1

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Answer 2

sin2x-sin4x=0
sin2x - 2 sin2x*cos2x=0
sin2x (1-2 cos2x) =0
sin 2x =0 ⇒2x=πn, n∈Z ⇒ x₁ = πn 2, n∈Z
или
1-2 cos2x=0⇒cos2x=12 ⇒2x=⁺- arccos 12+πn, n∈Z; 2x=⁺-π3+2πn, n∈Z
x₂= ⁺-π6+πn, n∈Z