Question

已知向量a=(√3sinx,cosx),b=(cosx,cosx)函数f(x)=2ab-1

(1)求f(x)的最小正周期(2)当x∈[π/6,π/2]时,若f(x)=1,求x的值 速度,,快的多加分........

Answer 1

ab=√3*sinx*cosx+cos^2x.
=1/2(2sinx*cosx*√3+2cos^2x)
=1/2(sin2x*√3+cos2x+1)
=sin2x*1/2+cos2x*√3/2+1/2
=sin(2x+π/6)+1/2.

f(x)=2ab-1=2sin(2x+π/6)+1-1
=2sin(2x+π/6).

T=2π/2=π.
则,f(x)的最小正周期是:π.

因为f(x)=1,则有
2sin(2x+π/6)=1,
sin(2x+π/6)=1/2.
而,x∈[π/6,π/2],则有
π/6≤X≤π/2,
π/3≤2X≤π,
π/3+π/6≤2X+π/6≤π+π/6,
即,π/2≤2X+π/6≤7π/6.
sin(2x+π/6)=1/2.所以有,
2X+π/6=5π/6,
X=π/3.

Answer 2

（1）周期π

最后化出的式子是：f(x)=2sin(2x+π/6)

Answer 3

1
f(x)=2ab-1
=2(√3sinx·cosx+cos^2 x)-1
=√3sin2x+cos2x
=2·[(√3/2)·sin2x+(1/2)·cos2x]
=2·[cos(π/6)·sin2x+sin(π/6)·cos2x]
=2·sin(2x+π/6)
∴最小正周期是2π/2=π.

2
当x∈[π/6,π/2]时,2x+π/6∈[π/2,7π/6]
若
f(x)=1
即2·sin(2x+π/6)=1
则sin(2x+π/6)=1/2
则2x+π/6=5/6
x=π/3