Sin(2x+2pi/3)cos(4x+pi/3)-cos2x=sin^2(x)/cos(-pi/3) - вопрос №2234322315557694 от razebal 21.07.2020 19:00

Question

Алгебра: Sin(2x+2pi/3)cos(4x+pi/3)-cos2x=sin^2(x)/cos(-pi/3)

razebal · Answer

Пошаговое объяснение:Sin(2x+2pi/ 3)cos(4x+pi/3)-cos2x-sin ^2(x)/cos(-pi/3), Sin(2x+2pi/3)cos(4x*pi/3) %3D 2 sin ^2(x)+ cos2x, Sin(2x+2pi/3) cos(4xtpi/3) - 1 1) (Sin(2x+2pi/3 )-1 (cos(4xtpi/3) %3 1 (2x+2pi/3- pi/2 +2pin, х %3- pi/12 +pin. n e (4xtpi/3-2pin, х3- pi/12 +pin/ 2%3D x%3- pi/12 + pin. n € Z 2) (Sin(2x+2pi/3)- -1 (cos(4x+pi/3)-D- 1 (2x+2pi/3-- pi/2 +2pik, х %3D- Трi/12+pik,k € z (4x*pi/3 %3D pi +2pik, х3D pi/6 +pik/ 2,k € Z %3D...