a·b=cos3/2xcosx/2-sin3/2xsinx/2=cos2x
|a+b|=|(cos3/2x,sin3/2x)+(cosx/2,-sinx/2)|
=|(cos3/2x+cosx/2,sin3/2x-sinx/2)|
=√[(cos3/2x+cosx/2)^2+(sin3/2x-sinx/2)^2]
=√[2+2(cos3/2xcosx/2-sin3/2xsinx/2)]
=√[2+2cos2x]=√[2(1+cos2x)]=2cosx(x∈【-π/3,π/4】,cosx>0)
所以：f(x)=cos2x-2cosx=2cos^x-2cosx-1
=2[cosx-1/2]^2-3/2
所以,当cosx=1/2,即x=-π/3时f(x)取最小值-3/2;
由于x∈【-π/3,π/4】,cosx在此区间的最大、最小值为1,1/2;
所以：当cosx=1,即x=0时,f(x)取最大值-1.