高一 数学 关于函数。 请详细解答,谢谢! (13 16:19:35)

cos2x-sin2x≠0
tan2x≠1
2x≠kπ+π/4
x≠kπ/2+π/8

f(x)=cos4x/cos2x-sin2x
=（cos^2 2x-sin^2 2x)/(cos2x-sin2x)
=(cos2x+sin2x)(cos2x-sin2x)/(cos2x-sin2x)
=cos2x+sin2x
=√2sin(2x+π/4)
1)
f(x)的定义域:x≠kπ/2+π/8,k∈Z
周期:2kπ/2=kπ，k∈Z
最小正周期：π

2）
单调递增区间：2x+π/4∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-3π/8,kπ+π/8],x≠kπ/2+π/8,k∈Z

f(x)=cos4x/(cos2x-sin2x)
=[(cos2x)^2-(sin2x)^2]/[cos2x-sin2x]=cos2x+sin2x
=√2[sin45cos2x+sin2xcos45]
=√2sin(45+2x)
=√2sin(Pai/4+2x)

cos2x-sin2x≠0
2x≠Pai/4
x≠Pai/8
所以定义域：x∈R且x≠Pai/8

周期:2Pai/2=Pai

单调递增区间:2kPai-Pai/2=<Pai/4+2x<=2kPai+Pai/2
2kPai-3Pai/4<=2x<=2kPai+Pai/4
kPai-3Pai/8<=x<=kPai+Pai/8
即区间是:[kPai-3Pai/8,kPai+Pai/8]