Sin2x=sin4x/2-cos4x/2 решить уравнение на отрезке [-п/2;п/2]

sin2x=(sin²x/2-cos²x/2)(sin²x/2+cos²x/2)sin2x=-cosx2sinxcosx+cosx=0cosx(2sinx+1)=0cosx=0⇒x=π/2+πk,k∈z-π/2≤π/2+πk≤π/2-1≤1+2k≤1-1≤k≤0k=-1  x=π/2-π=-π/2k=0  x=π/22sinx+1=0⇒sinx=-1/2⇒x=-π/6+2πk U x=-5π/6+2πk,k∈z-π/2≤-π/6+2πk≤π/2-3≤-1+12k≤3-1/6≤k≤1/3k=0  x=-π/6-π/2≤-5π/6+2πk≤π/2-3≤-5+12k≤31/6≤k≤2/3 нет решенияОтвет {π/2+πk;-π/6+2πk;-5π/6+2πk,k∈z};-π/2;π/2;-π/6