(1)f(x)=(1/2)(cos^2x-sin^2x)-(√3/2)(2sinxcosx)+1
=1/2cos2x-(√3/2)*sin2x+1
=cos(2x+π/3)+1
最小正周期为π,区间上最大值3/2,最小值1/2.
(2)cos(2x1+π/3)+1=9/5 x1∈[-π/6,π/6]
so (2x1+π/3)∈[0,2π/3]
so cos(2x1+π/3)=4/5 sin(2x1+π/3)=3/5
so cos2x1=cos(2x1+π/3-π/3)=1/2*4/5+√3/2*3/5=(4+3√3)/10