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

Sin2x=(sin²x/2-cos²x/2)(sin²x/2+cos²x/2)
sin2x=-cosx
2sinxcosx+cosx=0
cosx(2sinx+1)=0
cosx=0⇒x=π/2+πk,k∈z
-π/2≤π/2+πk≤π/2
-1≤1+2k≤1
-1≤k≤0
k=-1  x=π/2-π=-π/2
k=0  x=π/2
2sinx+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≤1/3
k=0  x=-π/6
-π/2≤-5π/6+2πk≤π/2
-3≤-5+12k≤3
1/6≤k≤2/3 нет решения
ответ {π/2+πk;-π/6+2πk;-5π/6+2πk,k∈z};-π/2;π/2;-π/6