求y=sin(2x+π/3)+cos(2x-π/6)的单调区间
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y=sin(2x+π/3)+cos(2x-π/6)
=(1/2)sin2x+(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x
=sin2x+√3cos2x
=2sin(2x+π/3)
2kπ-π/2<=2x+π/3<=2kπ+π/2，kπ-5π/12<=x<=kπ+π/12
2kπ+π/2<=2x+π/3<=2kπ+3π/2，kπ+π/12<=x<=kπ+7π/12
递增区间为：[kπ-5π/12,kπ+π/12]；递减区间为：[kπ+π/12,kπ+7π/12]
=(1/2)sin2x+(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x
=sin2x+√3cos2x
=2sin(2x+π/3)
2kπ-π/2<=2x+π/3<=2kπ+π/2，kπ-5π/12<=x<=kπ+π/12
2kπ+π/2<=2x+π/3<=2kπ+3π/2，kπ+π/12<=x<=kπ+7π/12
递增区间为：[kπ-5π/12,kπ+π/12]；递减区间为：[kπ+π/12,kπ+7π/12]
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