1-sin3x=(sinx/2-cosx/2)²

Question

1-sin3x=(sinx/2-cosx/2)²

Answer 1

1-sin3x=sin²x/2-2sinx/2*cosx/2+cos²x/2
1-sin3x=1-sinx
sin3x-sinx=0
2sinxcos2x=0
sinx=0⇒x=πn,n∈z
cos2x=0⇒2x=π/2+πk,k∈z⇒x=π/4+πk/2,k∈z