f(x)=sinx^4+2倍根号3 sinxcosx-cosx^4 求（1）函数f(x)最小正周期 （2）函数在闭区间0

(1)f(x)=sinx^4+2√3sinxcosx-cosx^4
=sinx^4-cosx^4+√3sin2x
=√3sin2x+(sinx^2+cosx^2)(sinx^2-cosx^2)
=√3sin2x-cos2x
=2[sin2x*(√3/2)-cos2x*(1/2)]
=2[sin2x*cos(π/6)-cos2x*sin(π/6)]
=2sin(2x-π/6)
∴f(x)最小正周期T=2π/ω=2π/2=π.
(2)令2x-π/6∈[2kπ-π/2,2kπ+π/2](k∈Z),得x∈[kπ-π/6,kπ+π/3](k∈Z),
则f(x)在区间[0,π/3]和[5π/6,π]上是增函数
令2x-π/6∈[2kπ+π/2,2kπ+3π/2](k∈Z),得x∈[kπ+π/3,kπ+5π/6](k∈Z),
则f(x)在区间[π/3,5π/6]上是减函数.

1.f(x)=(sinx^2-cosx^2)+2倍根号3 sinxcosx
=2倍根号3 sinxcosx-cos2x
=根号3 sin2x-cos2x
=2sin(4x-π/6)
函数f(x)最小正周期为π/4
递增区间 为：4x-π/6小于或等于π/2大于或等于0
既 区间为： π/24和π/6之间
辛苦啊 ·· 好难打字的。。

(1)f(x)=sinx^4+2√3 sinxcosx-cosx^4
=sinx^4-cosx^4+√3 sin2x
=(sinx^2+cosx^2)(sinx^2-cosx^2)+√3 sin2x
=-cos2x+)+√3 sin2x
=2sin(2x-π/6)
所以f(x)最小正周期为π.
(2)由所以2x-π/6∈[2kπ-π/2,2kπ+π/...