已知函数fx=sin(2x+π/6)+sin(2x-π/6)+2cos^2x(x属于R)
1.求函数fx的最大值及此时自变量函数x的取值集合2.求函数fx的单调递增区间3.求使fx≥2x的x的取值范围
(1)解析:∵函数f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos^2x
=√3sin2x+cos2x+1=2sin(2x+π/6)+1
2x+π/6=2kπ+π/2==>x=kπ+π/6
∴当x=kπ+π/6时，函数fx的最大值为3
(2)解析:2kπ-π/2<=2x+π/6=2kπ+π/2==>kπ-π/3<=x=kπ+π/6
∴函数fx的单调递增区间为kπ-π/3<=x=kπ+π/6
(3)解析:∵f(x)≥2x
设g(x)=
2sin(2x+π/6)+1-2x>=0
解得x≈1.02981
∴当x<=1.02981时，满足f(x)≥2x
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