已知函数f(x)=2cos(kx+π/3)+b(k>0)的最小正周期为π，最小值为3

已知函数f(x)=2cos(kx+π/3)+b(k>0)的最小正周期为π，最小值为3
（1）求k,b的值，并写出相应的f(x)解析式

k=2,b=5
f(x)=2cos(2x+π/3)+5

（2）求M(x)=F(x-π/4)+F(x)的最大值和最小值

M(x)=F(x-π/4)+F(x)=2cos(2(x-π/4)+π/3)+5+2cos(2x+π/3)+5

=2(cos(2x-π/2+π/3)+2cos(2x+π/3))+10
=2(sin(2x+π/3)+cos(2x+π/3))+10
=2√2sin(2x+π/3+π/4)+10
sin(2x+π/3+π/4)的最大值1,最小值-1

M(x)的最大值10+2√2,最小值10-2√2