Question

已知函数f(x)=cos^2x-sin^2x+2根号3sinxcosx+1

求f(x)最小正周期和单调递增区间.若f（a）=2.a属于[派/4,派/2]求a值

Answer 1

f(x)=cos^2x-sin^2x+2(根号3)sinxcosx+1
=cos2x+(根号3)sin2x+1
=2{(1/2)cos2x+[(根号3)/2]sin2x}+1
=2sin(2x+派/6)+1,
T=2派/2=派
递增区间[K派-派/3,K派+派/6]
2sin(2a+派/6)+1=2,
2a+派/6=5派/6,a=派/3,