Question

Cosx/2*cosx*cos2x*cos4x=1/16 решите подробно.

Answer 1

Cosx/2*cosx*cos2x*cos4x=1/16
умножим и разделим левую часть на sinx/2
(sinx/2*Cosx/2*cosx*cos2x*cos4x)/(sinx/2)=
=1/2*(sinxcosx*cos2x*cos4x)/(sinx/2)=1/4*(sin2x*cos2x*cos4x)/(sinx/2)=
=1/8*(sin4x*cos4x)sin(x/2)=1/16*sin8x/(sinx/2)
1/16*sin8x/(sinx/2)=1/16
sin8x/(sinx/2)=1
sin8x=sinx/2,sinx/2≠0
sin8x-sinx/2=0
2sin(15x/4)*cos(17x/4)=0
sin(15x/4)=0⇒15x/4=πk⇒x=4πk/15,k∈z
cos(17x/4)=0⇒17x/4=π/2+πk⇒x=2π/15+4πk/15,k∈z