f(x)＝6sinxcosx＋4cos(x＋π/4)cos(x－π)＋2cos^2x
＝3sin(2x)＋2√2(cosx－sinx)(﹣cosx)＋2cos^2x
＝3sin(2x)＋2√2sinxcosx－2√2cos^2x＋2cos^2x
＝(3＋√2)sin(2x)－(√2－1)[cos(2x)＋1]
＝√(14＋4√2)sin(2x－θ)－(√2－1) tanθ＝(√2－1)/(3＋√2)＝(4√2－5)/7
∴f(x)的最小正周期T＝2π/2＝π
当2x－θ＝2kπ＋π/2即x＝kπ＋π/4＋θ时,取得最大值√(14＋4√2)－(√2－1)