((sin^3(x)*cos3x)/cosx)+cos^2(x)*sin3x=6*cos2x*sin^2(x) - вопрос №58025083323622 от morgo19 24.12.2022 14:20

Question

Математика: ((sin^3(x)*cos3x)/cosx)+cos^2(x)*sin3x=6*cos2x*sin^2(x)

morgo19 · Answer

Левая часть. Разность кубовsin^3 x - cos^3 x = (sin x - cos x)(sin^2 x + sin x*cos x + cos^2 x)Правая часть1 + sin 2x / 2 = sin^2 x + cos^2 x + sin x*cos xПолучаем(sin x-cos x)(sin^2 x+sin x*cos x+cos^2 x) = sin^2 x+cos^2 x+sin x*cos x(sin^2 x + cos^2 x + sin x*cos x)(sin x - cos x - 1) = 01) sin^2 x + cos^2 x + sin x*cos x = 0 1 + sin 2x / 2 = 0sin 2x = -2 - решений нет2) sin x - cos x - 1 = 02sin(x/2)*cos(x/2) - cos^2(x/2) + sin^2(x/2) - cos^2(x/2) - sin^2(x/2) = 02sin(x/2)*cos(x/2) - 2cos^2(x/2) = 02cos(x/2)*(sin(x/2) - cos(x/2)) = 0cos(x/2) = 0; x/2 = pi/2 + pi*k; x1 = pi + 2pi*k
sin(x/2) - cos(x/2) = 0sin(x/2) = cos(x/2)tg(x/2) = 1; x/2 = pi/4 + pi*k; x2 = pi/2 + 2pi*k