Question

1.Determinati numerele reale x∈(0,π)pentru care sin2x=sinx

Answer 1

sin 2x=sinx=>sin2x-sinx=0=>2sin[(2x-x)/2]cos[(2x+x)/2]=0=>2sin(x/2)cos(3x/2)=0=>
sin(x/2)=0=>x/2∈{kπ|k∈Z}=>x∈(2kπ|k∈Z}
cos(3x/2)=0=>3x/2∈{π/2+kπ|k∈Z}=>3x∈{π+2kπ|k∈Z}=>x∈{π/3+2kπ/3|k∈Z}