急！已知函数fx=2cosx(sinx-√3cosx)+√3,x属于R 求最小正周期和单调

求最小正周期和单调递增区间

f(x)=2sinxcosx-2√3(cosx)^2+√3
=sin2x-2√3*(cos2x+1)/2+√3
=sin2x+√3cos2x
=2sin(2x+π/3)
所以T=2π/2=π
2x+π/3属于[2kπ-2/π,2kπ+2/π]
x属于[kπ-5π/12,kπ+π/12]时单调递增

f(x) = sinx + (√3)cosx

= 2[(1/2)sinx + (√3/2)cosx]

= 2sin(x + π/3)

最小正周期为t = 2π/1 = 2π

x + π/3 = π/2 + 2kπ【k∈z】

x = π/6 + 2kπ

最大值为f(x)max = f(2kπ + π/6) = 2×1 = 2

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