已知函数f（x）=sin(2x-π/6)+2cos^2x-1 求单调区间
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f(x)=sin(2x-π/6)+2cos²x-1
=sin(2x-π/6)+cos2x
=sin2x×根号下3/2-cos2x/2+cos2x
=sin2x×根号下3/2+cos2x/2
=sin(2x+π/6)
单调增区间：2kπ-π/2<=2x+π/6<=2kπ+π/2
kπ-π/3<=x<=kπ+π/6
单调减区间：2kπ+π/2<=2x+π/6<=2kπ+3π/2
kπ+π/6<=kπ+2π/3
=sin(2x-π/6)+cos2x
=sin2x×根号下3/2-cos2x/2+cos2x
=sin2x×根号下3/2+cos2x/2
=sin(2x+π/6)
单调增区间：2kπ-π/2<=2x+π/6<=2kπ+π/2
kπ-π/3<=x<=kπ+π/6
单调减区间：2kπ+π/2<=2x+π/6<=2kπ+3π/2
kπ+π/6<=kπ+2π/3
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