题目

已知直线y=1与函数f(x)=2cos^2wx/2+cos(wx+π/3)
w大于0,其图像交于M,N,且MN绝对值等于π/2,求f(x单调减区间)

提问时间：2021-04-06

答案

f(x)=2cos^2wx/2+cos(wx+π/3)
=(1+coswx)+cos(wx+π/3)
=1+coswx+coswxcos(π/3)-sinwxsin(π/3)
=1+(3/2)coswx-(√3/2)sinwx
=1+√3[coswxcos(π/6)-sinwxsin(π/6)]
=1+√3cos(wx+π/6)
y=1与其交于两点M,N,则|MN|=T/2=π/2
所以 T= π=2π/w
w=2
所以 f(x)=1+√3cos(2x+π/6)
求减区间 即 2kπ≤2x≤2kπ+π
2kπ-π/6≤2x≤2kπ+5π/6
kπ-π/12≤x≤kπ+5π/12
减区间为[ kπ-π/12,kπ+5π/12],k∈Z