高一数学!!!!!急!!!!!
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f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos²x
=sin2xcosπ/6+sinπ/6cos2x+sin2xcosπ/6-sinπ/6cos2x+2cos²x
=2sin2xcosπ/6+2cos²x
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
∴当2x+π/6=2kπ+π/2即x=kπ+π/6,k∈Z时,f(x)的最大值是3
(2)
由2x+π/6∈[2kπ-π/2,2kπ+π/2],k∈Z可知:单增区间是
[kπ-π/3,kπ+π/6],k∈Z
=sin2xcosπ/6+sinπ/6cos2x+sin2xcosπ/6-sinπ/6cos2x+2cos²x
=2sin2xcosπ/6+2cos²x
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
∴当2x+π/6=2kπ+π/2即x=kπ+π/6,k∈Z时,f(x)的最大值是3
(2)
由2x+π/6∈[2kπ-π/2,2kπ+π/2],k∈Z可知:单增区间是
[kπ-π/3,kπ+π/6],k∈Z
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