2 вопроса:
1. Как сделали последние преобразование не пойму: cos2x =cos²x - sin²x =cos²x -(1 -cos²x) =2cos²x -1⇒cos²x =(1+cos2x)/2
2. Правильно ли решено уравнение?
sin4x= sin2x * * * sin4x =sin2*2x = 2sin2x*cos2x * * *
2sin2x*cos2x =sin2x ;
2sin2x*cos2x - sin2x =0 ;
2sin2x(cos2x - 1/2) =0 ;
[ sin2x = 0 ; [ 2x =πk , k ∈ Z ;
[cos2x =1/2 . [ 2x =±π/3 +2πn , n∈Z.

[ x =(π/2)*k , k ∈ Z ;
[ x =±π/6 +πn , n∈Z.
Просто в ответе: x=pi n и x=+- pi/6+pi/3n
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