已知函数f（x）=sinxcosx＋cos的平方x（x∈R）.（1）求函数f（x）的最小正周期,

f(x)=sinxcosx+cosx^2=(1/2)*2sinxcosx+(1/2)*(2cosx^2-1)+1/2
=(1/2)*sin2x+(1/2)*cos2x+1/2
=(√2/2)*(√2/2sin2x+√2/2cos2x)+1/2
=(√2/2)*sin(2x+π/4)+1/2
(1)T=2π/ω=π
(2)f(x)=(√2/2)*sin(2x+π/4)+1/2
当sin(2x+π/4)=1时,f(x)取到最大值(1+√2)/2
此时2x+π/4=π/2+2kπ(k∈Z) -> x=π/8+kπ
故当f(x)取到最大值时,解集为{x|x=π/8+kπ,k∈Z}

f(x)=sinxcosx+cos²x
=(1/2)sin2x+(1+cos2x)/2
=(1/2)sin2x+(1/2)cos2x+1/2
=(√2/2)sin(2x+π/4)+1/2
(1)
函数f(x)的最小正周期是T=2π/2=π
(2)
函数f（x）的最大值是√2/2+1/2=(√2+1)/2
此时2x+π/4=2...