Cos^2x+cos^2*2x+cos^2*3x+cos^2*4x=2 - вопрос №226132822223242 от emkaemkovich902 07.06.2020 07:22

Question

Алгебра: Cos^2x+cos^2*2x+cos^2*3x+cos^2*4x=2

emkaemkovich902 · Answer

(1+cos2x)/2+(1+cos4x)/2+(1+cos6x)/2+(1+cos8x)/2=4
4+cos2x+cos4x+cos6x+cos8x=4
(cos2x+cos8x)+(cos4x+cos6x)=0
2cos5xcos3x+2cos5xcosx=0
2cos5x(cos3x+cosx)=0
2cos5x*2cos2xcosx=0
cos5x=0⇒5x=π/2+πn⇒x=π/10+πn/5
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2
cosx==0⇒x=π/2+πn