1) sinx+sin2x+sin3x+sin4x=0 2) cosx+cos2x+cos3x+cos4x=0 

1) sinx+sin2x+sin3x+sin4x=0

2) cosx+cos2x+cos3x+cos4x=0 

Ответ:

1)sinx+sin2x+sin3x+sin4x = (sinx+sin3x)+(sin2x+sin4x) =2sin2x× cosx+2sin3x× cosx =2cosx(sin2x+sin3x) =4sin(5x/2)× cos(x/2)× cosx

sin(5x/2)=0 5x/2=p n x=2p n/5

cos(x/2)=0 => x/2=p /2+p n э Z => x=p +2p n

cosx=0 x=p /2+p n x=p /2+p n

2p n/5 ; p (1+2n) ; p (1/2+n), n э Z.

2)cosx+cos2x+cos3x+cos4x=0

2cos(5/2)x*cos(x/2)*2*cos(5/2)x*cos(3/2)x=0

cos(5/2)x(cos(x/2)+cos(3/2)x)=0

2cos(5/2)x*cosx*cos(x/2)=0

cos(5/2)x=0x=p/2 + 2/5 np, n э Z

cosx=0x=p/2+ ap, a э Z

cos(x/2)=0x=p+2bp, b э z

вроде так)

Источник: https://znanija.com/task/245365