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