已知函数fx =2sin(x-6分之派 )cosx+2cos平方x

题目

已知函数fx =2sin(x-6分之派 )cosx+2cos平方x
(1)求fx单调递增区间

答案

f(x) =2sin(x-π/6)cosx+2cos²x=2(sinxcosπ/6-cosxsinπ/6)cosx+2cos²x=√3sinxcosx+(cosx)^2=(√3/2)*sin2x+(cos2x+1)/2=(√3/2)*sin2x+(1/2)*cos2x+1/2=sin(2x+π/6)+1/2单调递增区间：2kπ-π/2...