[tex]sin^{2}x+\frac{1}{2}.sin^{2}3x-sinx.sin^{2}3x=\frac{1-cos2x}{2}+\frac{1-cos6x}{4}-\frac{(1-cos6x)(1-cos2x)}{4}=4(3-2.cos2x-cos6x-1+cos6x+cos2x+cos6x.cos2x)=-4(cos2x+cos2x.cos6x-2)==0[/tex]
=> [tex](cos6x+1).cos2x=2[/tex]
Mà [tex](cos6x+1).cos2x\leq 2[/tex]
Dấu = xảy <=>cos6x=cos2x=1