已知函数f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx
1个回答

1.

f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx

= 2cosx*sin(x+π/3)- 2sinx*[(√3/2)sinx-(1/2)cosx]

= 2cosx*sin(x+π/3)- 2sinx*[sin(π/3)sinx-cos(π/3)cosx)]

= 2cosx*sin(x+π/3)+ 2sinxcos(x+π/3)

= 2sin(2x+π/3)

f(x)的单调区间

2kπ - π/2

2sin[2(x-m)+π/3] = 2sin{[2(x+m)-π/3]}

=>

2(x-m) + π/3 + 2kπ = 2(x+m)-π/3

=>

4m= 2π/3 + 2kπ

m=π/6 取最小正值(k=0)

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