Question

设函数f（x）=3sin2x+2√3cos²x+3-√3,x∈R,求f（x）的单调区间

Answer 1

f(x) = 3sin2x+2√3(cosx)^2+3-√3
= 3sin2x+√3(1+cos2x)+3-√3
= 3sin2x+√3cos2x+3
= 2√3sin(2x+π/6)+3
当 2kπ-π/2 ≤ 2x+π/6 ≤ 2kπ+π/2,
即 kπ-π/3 ≤ x ≤ kπ+π/6 时， 函数单调增加；
当 2kπ+π/2 ≤ 2x+π/6 ≤ 2kπ+3π/2,
即 kπ+π/6 ≤ x ≤ kπ+2π/3 时， 函数单调减少。