已知函数f(x)=2√3sinxcosx-2sin눀x+1

2024-11-30 05:36:21
推荐回答(1个)
回答1:

f(x)=2√3sinxcosx-2sin²x+1
=√3sin2x+cos2x
=2sin(2x+π/6)
(2) 2kπ-π/2<=2x+π/6<=2kπ+π/2
kπ-π/3<=x<=kπ+π/6
增区间 【kπ-π/3,kπ+π/6】 k∈Z
(2)x∈【0,π/2】
2x+π/6∈【π/6,7π/6】
x=π/2 fmin=-1
(3) f(x0)=6/5
sin(2x0+π/6)=3/5 x0∈[π/4,π/2],2x0+π/6∈[2π/3,7π/6],
cos(2x0+π/6)=-4/5
sin2x0=sin[(2x0+π/6)-π/6]
=sin(2x0+π/6)cosπ/6-cos(2x0+π/6)sinπ/6
=3/5*√3/2+4/5*1/2
=(3√3+4)/10