sin(x+π/4)=cos(π/2-x-π/4)=cos(π/4-x)=cos(x-π/4)
所以,2sin(x-π/4)sin(x+π/4)=2sin(x-π/4)cos(x-π/4)=sin(2x-π/2)=-cos2x
所以,f(x)=cos(2x-π/3)-cos2x=-2sin(2x-π/6)sin(-π/6) （这一步要用到和差化积公式）
所以,sin(2x-π/6)=0
所以,x=(k/2+1/12)π （其中k为整数）
已知函数f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4),求函数f(x)的零点.
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