Решить уравнение: cos2x=1-cos(п/2-х) и найти корни на промежутке [-5/2п; -п]

4*(1+cos2x)²/4  -1 =cos2x
(1+cos2x)² -(1+cos2x)=0
(1+cos2x)(1+cos2x-1)=0
(1+cos2x)=0⇒cos2x=-1⇒2x=π+2πn⇒x=π/2+πn⇒x=-π/2∈[-5π/6;π/6]
(1+cos2x-1)=0⇒cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2⇒x=-3π/4;-π/4∈[-5π/6;π/6]