Question

已知f（x）=2cosxsinx+2根号3cos平方x-根号3

求最小正周期和单调增区间求过程

Answer 1

f(x)=2cosxsinx+2√3cos^2x-√3
=sin2x+√3(1+cos2x)-√3
=sin2x+√3cos2x
=2(sin2x/2+√3cons2x/2)
=2sin(2x+π/3)
T=2π/ω=π
单调递增区间 - π/2+2kπ≤2x+π/3≤π/2+2kπ 解得: [-5π/12+kπ,π/12+kπ]
单调递减区间 π/2+2kπ ≤2x+π/3≤2π+2kπ 解得 [π/12+kπ,5π/6+kπ]

追问

对嘛