f(x)
=a.b-1
=(2sinx,cosx).(√3cosx,2cosx)-1
=2√3sinxcosx +2(cosx)^2-1
=√3sin2x+cos2x
=2sin(2x+π/6)
最小正周=π
最大值=2
when
2x+π/6 = π/2
x= π/6
x属于[-7π/12,5π/12]
f(x)单调区间
增加
-π/2<=2x+π/6<= π/2
-π/6 <= x <= π/6
减小
-7π/12<=x <= -π/6 or π/6 <=x<= 5π/12
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