已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos²x
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f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos²x
=(sin2xcosπ/6+cos2xsinπ/6)-(cos2xcosπ/3-sin2xsinπ/3)+(cos2x+1)
=√3/2sin2x+1/2cos2x-1/2cos2x+√3/2sin2x+cos2x+1
=√3sin2x+cos2x+1
=2(√3/2sin2x+1/2cos2x)+1
=2(sin2xcosπ/6+cos2xsinπ/6)+1
=2sin(2x+π/6)+1
所以(1)f(π/12)=2sinπ/3+1=√3+1.
(2)当2x+π/6=π/2+2kπ (k∈Z)时，即x=π/6+kπ ( k∈Z)时，
sin(2x+π/6)取得最大值1，从而f(x)取得最大值3
=(sin2xcosπ/6+cos2xsinπ/6)-(cos2xcosπ/3-sin2xsinπ/3)+(cos2x+1)
=√3/2sin2x+1/2cos2x-1/2cos2x+√3/2sin2x+cos2x+1
=√3sin2x+cos2x+1
=2(√3/2sin2x+1/2cos2x)+1
=2(sin2xcosπ/6+cos2xsinπ/6)+1
=2sin(2x+π/6)+1
所以(1)f(π/12)=2sinπ/3+1=√3+1.
(2)当2x+π/6=π/2+2kπ (k∈Z)时，即x=π/6+kπ ( k∈Z)时，
sin(2x+π/6)取得最大值1，从而f(x)取得最大值3
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