решите уравнение
1-cos 4x=sin 2x
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2sin²2x = sin2x ;
2sin2x(sin2x -1/2) =0 ⇔
[ sin2x =0 ; sin2x =1/2.
a)
sin2x =0 ;
2x = π*n ,n ∈
Z ;
x = π*n /2 ,n ∈ Z.
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b)
sin2x =1/2 ;
2x = ((-1)^n)*π/6 +π*n , n∈ Z;
x =( (-1)^n )*π/12 +(π/2)*n , n∈ Z .
ответ : π*n /2 , ( (-1)^n )*π/12 +(π/2)*n , n ∈ Z.