已知函数f(x)=2cosx∧2+2√3sinxcosx-1 (1)求函数f

解由f(x)=2cosx∧2+2√3sinxcosx-1
=2cosx∧2-1+2√3sinxcosx
=cos2x+√3sin2x
=2（√3/2sin2x+1/2cos2x）
=2sin（2x+π/6）
1函数的周期T=2π/2=π
对称轴方程为2x+π/6=kπ+π/2,k属于Z
即对称轴方程为x=kπ/2+π/6,k属于Z,
2当2kπ-π/2≤2x+π/6≤2kπ+π/2时,k属于Z,函数是增函数
即当kπ-π/3≤x≤kπ+π/6时,k属于Z,函数是增函数
即函数的增区间为[kπ-π/3,kπ+π/6]k属于Z
当2kπ+π/2≤2x+π/6≤2kπ+3π/2时,k属于Z,函数是减函数
即当kπ+π/6≤x≤kπ+2π/3时,k属于Z,函数是减函数
即函数的增区间为[kπ+π/6,kπ+2π/3]k属于Z.

f(x)=2cosx∧2+2√3sinxcosx-1 =2sin(2x-π/6)
(1) T=2π/2=π
2x-π/6=kπ+π/2
对称轴x=kπ/2+π/3
(2)
2kπ-π/2<=2x-π/6<=2kπ+π/2
kπ-π/6<=x<=kπ+π/3
所以在(kπ-π/6,kπ+π/3)上单调递增
2kπ+π/2<=2x-π...