Вопрос от пользователя: Меркурия
Решите уравнение tg2x+tg4x=tg5x+tgx
Ответ: Олюля
Tg2x+tg4x=tg5x+tgx
sin6x/cos2xcos4x=sin6x/cos5xcosx
cos2xcos4x=cos5xcosx
1/2 (cos2x+cos6x) = 1/2 (cos4x+cos6x)
cos2x+cos6x=cos4x+cos6x
cos2x=cos4x
2cos²2x-1-cos2x=0
cos2x=a
2a²-a-1=0
D=1+8=9
a1 = (1-3) / 4=-1/2⇒cos2x=-1/2⇒2x=+2π/3+2πn, n∈Z⇒x=+π/3+πn, n∈Z
a2 = (1+3) / 4=1⇒cos2x=1⇒2x=2πk, k∈Z⇒x=πk, k∈Z
