Решить тригонометрию 3cos5x-2(cos6x+cos4x)/3sin5x-2(sin6x+sin4x)=tg17x - вопрос №2923147352643526417 от vegyan0000 30.06.2020 14:14

Question

Алгебра: Решить тригонометрию 3cos5x-2(cos6x+cos4x)/3sin5x-2(sin6x+sin4x)=tg17x

vegyan0000 · Answer

[3cos5x-4cos5x*cosx]/[3sin5x-4sin5xcosx]=tg17x
cos5x(3-4cosx)/sin5x(3-4cosx)=tg17x
cosx≠3/4
ctg5x=tg17x
tg17x-tg(π/2-5x)=0
sin(22x-π/2)/cos17x*cos(π/2-5x)=0
cos17x≠0 U cos(π/2-5x)=sin5x≠0
sin(22x-π/2)=0
22x-π/2=πn
22x=π/2+πn
x=π/44+πn/22