Question

sinx+sin3x=sin2x+sin4x

Answer 1

sinx+sin3x-sin2x+sin4x=0
2sin2xcosx-2sin3xcosx=0
2cosx(sin2x-sin3x)=0
2cosx*(-2sinx/2*cos5x/2)=0
cosx=0⇒x=π/2+πn,n∈z
sinx/2=0⇒x/2=πk⇒x=2πk,k∈z
cos5x/2=0⇒5x/2=π/2+πm⇒x=π/5+2πm/5,m∈z