Question

sin2x+sin4x+sin6x=0​

Answer 1

Răspuns:

x∈Kπ ∪(kπ+-π/3), k∈Z

Explicație pas cu pas:

fie 2x=y

siny+sin2y+sin3y=siny+sin3y+sin2y=2sin2ycosy+sin2y

am tunt con ca siny+sin3y = sin ((y+3y)/2) +cos((y-3y)/2) si ca, cos y e functie para

2sin2ycosy+sin2y=0

sin2y(2cosy+1)=0

sin2y=0

y=kπ.. x=kπ/2, k∈Z

2cosy+1=0

cosy=-1/2

y=2kπ+-arccos(-1/2) =2kπ+-2π/3

x=kπ+-π/3

x∈Kπ ∪(kπ+-π/3), k∈Z