f(x)=2cos(x+π/6)-1+1+√3sin2x
=cos(2x+π/3)+√3sin2x+1=
=[(1/2)cos2x-(√3/2)sin2x]+)+√3sin2x+1
=[(√3/2)sin2x＋(1/2)cos2x]+1
=sin(2x+π/6)+1
f(最大值)=2
f(最小值)= 0
將 2x+π/6代入到標(biāo)準(zhǔn)函數(shù)的單調(diào)增區(qū)間中去解出該函數(shù)的單調(diào)增區(qū)間如下 ：
－π/2+2kπ≤2x+π/6≤π/2+2kπ
－π/3+kπ≤x≤π/6+kπ
原函數(shù)的單調(diào)增區(qū)間為：【－π/3+kπ,π/6+kπ】
將 2x+π/6代入到標(biāo)準(zhǔn)函數(shù)的單調(diào)減區(qū)間中去解出該函數(shù)的單調(diào)減區(qū)間如下 ：
π/2+2kπ≤2x+π/6≤3π/2+2kπ
π/6+kπ≤x≤2π/3+kπ
原函數(shù)的單調(diào)減區(qū)間為：【π/6+kπ,2π/3+kπ】