已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x,求f(x)的最大值及相应x的值.

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已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x,求f(x)的最大值及相应x的值.
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已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x,求f(x)的最大值及相应x的值.
已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x,求f(x)的最大值及相应x的值.

已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x,求f(x)的最大值及相应x的值.
f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x      ( 
=(sin2x*cosπ/6 + cos2x*sinπ/6 ) - ( cos2x*cosπ/6 + sin2x*sinπ/6 )+ ( cos2x +1)
上传图片,

f(x)=sin(2x+π/6)-sin(π/6-2x)+cos2x+1
=2cos(π/6)sin2x+cos2x+1
=√2sin(2x+π/4)+1
当2x+π/4=2kπ+π/2, x=kπ+π/8时
f(x)有最大值=1+√2

解:sin(2x+π/6)+cos(2x+π/3)
=√2sin(2x+π/6+π/4)
=√2sin(2x+5π/12) （公式AsinX+BcosX=√(A^2+B^2) sin(X+ARGTAN（B/A）
T=2π/2=π,
2x+5π/12=π/2+2kπ时有最大值,
x=1/24π+kπ，f(x)=√2

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