Question

已知函数f(x)=2根号3sinXcosX+2cos^2X-1(X属于R)

1)求函数f(x)的最小正周期及在区间[0，派/2]上的最大值和最小值；2)若f(x0)=6/5，x0属于[派/4，派/2]，求cos2x0的值。

Answer 1

f(x)=根号3sin2x+cos2x=2sin(2x+π/6)
1)
T=2π/2=π
x∈[0,π/2]
2x+π/6[π/举让6,7π/6]
f(x)小=2sin(π/6)=1
f(x)大=2sin(π/2)=2
2)
sin(2x0+π/6)=3/5
co2x0=cos(2x0+π/6-π/宽答配6)=(3-4根慎指号3)/10
字限制

Answer 2

1) f(x)=2sin(2x+π/6)
最小正周期T=2π/2=π
因蔽行x∈[0,π/2]
故当x=π/6，f(x)最大=2，当x=π时，f(x)最小=-1
2）由 x0∈[π/4，宏悉哗π/2] 得陆配 cos2x0<0
√3sin2x0+cos2x0=6/5
∴cos2x0=(6-5√3)/20

Answer 3

f(x)=2根号3sinXcosX+2cos^2X-1
=根号3sin2X+2（1-sin^2X）-1
=-2sin^2X+3sin2X+1
=-2(sin2X-3/4)^+17/16
最大 当sin2x-3/4=0 时搜亏知空搜 f(x)=17/16
最小 当sin2x=-1 sin2x-3/4=-7/世消4 时 f(x)=-81/16