4sin²x +sin4x+2sin2x*sin4x =2;
4(1-cos2x)/2 +sin4x+ (cos2x -cos6x) =2;
2 - 2cos2x + sin4x+cos2x -cos6x =2 ;
sin4x -(cos6x+cos2x) =0;
2sin2x*cos2x+2cos4x*cos2x=0 ;
2cos2x(sin2x +cos4x) =0 ;
cos2x =0 ⇒2x =2πk ⇔ x₁=π*k k∈Z ;
sin2x +cos4x =0;
sin2x +(1 -2sin²x) =0;
2sin²x -sinx -1=0 ⇒sinx₂=1 ;sinx₂= -1/2 ;
x₂=π/2 +2π*k ;
x₃= (-1)^(k+1) π/6 +πk .
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ответ : π*k ; π/2 +2π*k ;(-1)^(k+1) π/6 +πk .