Question

已知函数f(x)=2根号3sinx*cosx+(cosx)^2-(sinx)^2-1(x属于R)

求函数y=f(x)的单调递增区间， 若x属于［-5丌/12，丌/3］，求f(x)的取值范围

Answer 1

f(x)=√3sin2x+cos2x-1=2sin(2x+π/6)-1
单调增区间为： 2kπ-π/2=<2x+π/6<=2kπ+π/2
即kπ-π/3=<x<=kπ+π/6, 这里k为任意整数。
若x属于[-5π/12, π/3],
则最大值为当x=π/6时取得，fmax=2-1=1
最小值为当当x=-π/3时取得，fmin=-2-1=-3
f(x)的范围是[-3,1]

Answer 2

f(x)=sqrt(3)sin2x+cos2x-1
f(x)=2sin(2x+pi/6)-1
2kpi-pi/2<=2x+pi/6<=2kpi+pi/2
2kpi-2pi/3<=2x<=2kpi+pi/3
kpi-pi/3<=x<=kpi+pi/6

x[-5pi/12,pi/3]
2x+pi/6[-2pi/3,5pi/6]
-1<=sin(2x+pi/6)<=1
-sqrt(3)-1<=f(x)<=sqrt(3)-1