Найдите в градусах сумму корней уравнения sin 2 (x) + sin 6 (x) = 2sin^2 (x) - 2cos^2 (x), принадлежащих отрезку [0; pi]

2sin4xcos2x=-2cos2x

2cos2x) sin4x+1) = 0

cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2

0≤π/4+πn/2≤π

-π/4≤πn≤3π/4

-1/4≤n≤3/4

n=0 x=π/4

sin4x=-1⇒4x=-π/2+2πn⇒x=-π/8+πn/2

0≤π/8+πn/2≤π

-π/8≤πn≤7π/8

-1/8≤n≤7/8

n=0 x=π/8

π/4+ππ/8=2π/8+π/8=3π/8=3*180/8=67 гр 30 мин