已知函数f(x)=4coswx·sin(wx-派/6)加1,求fx的单调递增区间

Question

已知函数f(x)=4coswx·sin(wx-派/6)加1,求fx的单调递增区间

Answer 1

f(x) = 4coswx·sin(wx-π/6) +1
= 4coswx·(sinwxcosπ/6-coswxsinπ/6) +1
= 4coswx·(√3/2sinwx-1/2coswx) +1
= 2√3sinwxcoswx - 2cos²wx +1
= √3sin2wx - cos2wx
= 2 { sin2wxcosπ/6 - cos2wxsinπ/6 }
= 2sin(2wx-π/6)
2wx-π/6属于（2kπ-π/2,2kπ+π/2）时单调增
单调增区间：x属于（kπ/w-π/(6w),kπ/w+π/(3w) )