Question

已知函数f（x)=2cosxsinx+2倍根号3cos^x-根号3 求f(x)的最小正周期 求f(x)的单调区间

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]