Question

已知f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx.

（1）求函数f(x)的最小正周期；
（2）当x∈[π/2,π]，函数f(x)=0的根。

Answer 1

f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx=
=[cos(3x/2+x/2)]-sin2x
=cos2x-sin2x
=√2[√2/2cos2x-√2/2sin2x]
=√2cos(2x+π/4)
(1)
最小正周期为T=2π/2=π
(2)π/2≤x≤π
5π/4≤2x+π/4≤2π+π/4
f(x)=0可化为；
cos(2x+π/4)=0
2x+π/4=3π/2;
x=5π/8

Answer 2

f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx.
=cos(3x/2+x/2)-sin2x
=cos2x-sin2x
=√2cos(2x+π/4)
(1)函数f(x)的最小正周期是2π/2=π
(2) f(x)=0 2x+π/4=kπ+π/2 x=(kπ+π/4)/2
π/2<=(kπ+π/4)/2<=π
π<=kπ+π/4<=2π 3/4<=k<=7/4 k=1 x=5π/8
当x∈[π/2,π]，函数f(x)=0的根是：x=5π/8

Answer 3

f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx=cos(3x/2+x/2)-sin2x
=cos2x-sin2x=根号2cos(2x+π/4)
所以f（x)的最小正周期=2π/2=π;
x∈[π/2,π]，2x∈[π,2π]，2x+π/4∈[5π/4,9π/4]，
f(x)=0; cos(2x+π/4)=0;
2x+π/4=3π/2
x=5π/8

Answer 4

f(x)=cos(3x/2+x/2)-sin2x
=cos2x-sin2x
=√2(√2/2cos2x-√2/2sin2x)
=√2(cos(π/4+2x))
(1) T=2π/2=π
(2) x∈[π/2,π] π=<2x<2π 5π/4=<π/4+2x<=9π/4
f(x)=√2cos(π/4+2x)=0
π/4+2x=3π/2
x=5π/8

Answer 5

f(x)=cos(3x/2+x/2)=cos2x
∴最小正周期是2π/2=π；在[π/2，π]上，若f(x)=0，则x=π/4

追问

可以详细点吗？

追答

f(x)=cos(3x/2+x/2)-2sinxcosx
=cos2x-sin2x
=√2cos(2x+π/4)
∴最小正周期是2π/2=π；
x∈[π/2，π]上，∴2x+π/4∈[5π/4，9π/4]，∴使得f(x)=0的点是2x+π/4=3π/2，解得x=5π/8