Question

已知函数f(x)=5sinxcosx-5根3cos^2x+5根3/2，（X∈R)，求f(x)的单调区间

Answer 1

f(x)=5/2*sin2x-5√3*(1+cos2x)/2+5√3/2
=5/2*sin2x-5√3/2*cos2x
=5(sin2x*1/2-cos2x*√3/2)
=5(sin2xcosπ/3-cos2xsinπ/3)
=5sin(2x-π/3)
sinx增则2kπ-π/2<x<2kπ+π/2
减则2kπ+π/2<x<2kπ+3π/2

所以单调增加则
2kπ-π/2<2x+π/6<2kπ+π/2
2kπ-2π/3<2x<2kπ+π/3
kπ-π/3<x<kπ+π/6
所以增区间(kπ-π/3,kπ+π/6)
同理，减区间(kπ+π/6,kπ+2π/3)