sin(2x)=sin(5x)<=>sin(2x)-sin(5x)=0<=>2sin(-3x/2)cos(7x/2)=0=> - sin(3x/2)cos(7x/2)=0
sin(3x/2)=0 sau cos(7x/2)=0
pt. sin(3x/2)=0=>3x/2={(-1)^k*arcsin(0)+kπ,k∈Z}=>x={2kπ/3,k∈Z}
pt. cos(7x/2)=0=>7x/2={+-arccos(0)+2kπ,k∈Z}=>x={+-2π/14+4kπ/7,k∈Z}=>x=+-π/7+4kπ/7,k∈Z}
Sper sa fie corect!