Question

已知函数f（x）=2√3sinxcosx+2cos²x-1（x∈R）

1.求函数f（x)的最小正周期及在区间[0，π/2]的最大和最小值 2.若f（x0）=6/5，x0∈[π/4,π/2],求cos2x0的值

Answer 1

f（x）=2√3sinxcosx+2cos²x-1（x∈R）
f(x) = 2√3sinxcosx+2cos²x-1
= √3 * ( 2sinxcosx) + (2cos²x-1)
= 根号3 sin2x + cos2x
= 2 (sin2xcosπ/6+cos2xsinπ/6)
=2 sin(2x+π/6)

最小正周期：2π/2 = π

在区间[0，π/2]
x ∈[0，π/2]
2x ∈[0，π]
2x+π/6 ∈[π/6，π+π/6]
2x+π/6∈[π/6，π/2]时单调增；
2x+π/6∈[π/2，π+π/6]时单调减
2x+π/6=π/2时有最大值，2sinπ/2=2
2x+π/6=π/6时，f(x)=2sinπ/6=1
2x+π/6=π+π/6时，f(x)=2sin(π+π/6)=-1
所以最小值-1，最大值2

f(x0）=6/5
2 sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5......(1)
x0∈[π/4,π/2]
2x0∈[π/2,π]
2x0+π/6 ∈[2π/3，7π/6]
cos(2x0+π/6) = - 根号{-sin^2(0+π/6)} = - {1-(3/5)^2} = -4/5......(2)

由（1）：sin2x0cosπ/6+cos2x0sinπ/6=3/5
根号3/2 sin2x0 + 1/2 cos2x0 = 3/5
根号3 sin2x0 + cos2x0 = 6/5 ......(3)

由（2）：cos2x0cosπ/6 - sin2x0sinπ/6 = -4/5
根号3/2 cos2x0 - 1/2 sin2x0 = -4/5
根号3 cos2x0 - sin2x0 = -8/5 ......(4)

(3)+(4)*根号3得：
cos2x0+3cos2x0 = 6/5 -8根号3/5
4cos2x0= 6/5 -8根号3/5
cos2x0= 3/10 - 2根号3 /5

求cos2x0

Answer 2

1.f（x）=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2sin(2x+π/6)
所以最小正周期T=π，
当x∈[0，π/2]时，2x+π/6∈[π/6，7π/6]，∴sin(2x+π/6)∈[-1/2,1]
所以f(x)最大值为2，最小值-1.
2.f（x0）=6/5知，sin(2x0+π/6)=3/5,由x0∈[π/4,π/2],知2x0+π/6∈[2π/3,7π/6],所以
cos(2x0+π/6)=-4/5,所以cos2x0=cos[(2xo+π/6)-π/6]=-4/5×√3/2+3/5×1/2=(3-4√3)/10

Answer 3

原式=跟3sin2x+cos2x=2(gen3sin2x+1/2cos2x)=2sin(30+2x)后边的可以画图求解。。

Answer 4

f（x）=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x
=2[(√3/2)sin2x+0.5cos2x]
注意到：
sin 60=√3/2 cos 60=1/2
因此f（x）=2（sin 60sin2x+cos 60cos2x）=2cos（2x-60°）
函数f（x)的最小正周期 2π/2=π
x在区间[0，π/2] 2x-π/3属于【-π/3，2π/3】
最大 2 和最小值为-1