f(x)=cos4次方—2根3sinxcosx—sin4次方 1.求周期 2.x属于[0,二分之派,f(x)min及取得最小值的集合 3.递增区间

(cosx)^4-(sinx)^4=(cos²x+sin²x)(cos²x-sin²x)=1*cos2x
2sinxcosx=2sin2x
所以f(x)=-√3sin2x+cos2x
=-(√3sin2x-cos2x)
=-√(3+1)sin(2x-z)
=-2sin(2x-z)
tanz=1/√3,z=π/6
f(x)=-2sin(2x-π/6)
1、
T=2π/2=π
2、
0-π/6所以2x-π/6=π/2,x=π/3时
f(x)最小=-2*1=-2
此时x=π/3
3、
-2sin(2x-π/6)递增则sin(2x-π/6)递减
所以2π+π/22π+2π/3π+π/3所以递增区间(π+π/3,kπ+2π/3)

f(x)=(cosx)^4-(sinx)^4-2sinxcosx
=[(sinx)^2+(cosx)^2][(cosx)^2-(sinx)^2]-2sinxcosx
=1*[(cosx)^2-(sinx)^2]-2sinxcosx
=cos2x-sin2x
=cos(2x+π/4)
所以最小正周期就是2π/2=π~