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
