f(x)=(1+cos2wx)/2+(√3/2)sin2wx
=sin2wx*cos(π/6)+cos2wx*sin(π/6)+1/2
=sin(2wx+π/6)+1/2
T =π=2π/2w
所以 w=1
f(x)=sin(2x+π/6)+1/2
(1)f(2π/3)=sin(4π/3+π/6)+1/2=sin(7π/6)+1/2=-1/2+1/2=0
(2) 增：
2kπ-π/2≤2x+π/6≤2kπ+π/2
2kπ-2π/3≤2x≤2kπ+π/3
kπ-π/3≤x≤kπ+π/6
增區(qū)間為 【kπ-π/3,kπ+π/6】,k∈Z
減：
2kπ+π/2≤2x+π/6≤2kπ+3π/2
2kπ+π/3≤2x≤2kπ+4π/3
kπ+π/6≤x≤kπ+2π/3
減區(qū)間為 【kπ+π/6,kπ+2π/3】,k∈Z
2x+π/6=kπ+π/2
2x=kπ+π/3
對稱軸方程 x=kπ/2+π/6,k∈Zf（2π/3）為什么我算的等于-1/2 是我錯了么？？我算錯了，抱歉 f(2π/3)=sin(4π/3+π/6)+1/2=sin(3π/2)+1/2=-1+1/2=-1/2