（sin^4 x+cos^4 x+sin^2 x * cos^2 x)/2-sin2x的最小正周期,最大值和最小值..

1、f(x)=(sin^4 x+cos^4 x+sin^2 xcos^2 x)/(2-sin2x)
=[（sin^2 x+cos^2 x)^2-2sin^2 xcos^2 x+sin^2 xcos^2 x]/(2-sin2x)
=(1-sin^2 xcos^2 x)/(2-sin2x)
=[(1-sinxcosx)(1+sinxcosx)]/[2(1-sinxcosx)]=(1+sinxcosx)/2
2、f(x)=4sinxsin^2(π/4 +x/2)+cos2x
=2sinx(cosx/2+sinx/2)^2+cos2x
=2sinx(1+sinx)+cos2x
=2sinx+2sin^2 x+1-2sin^2 x
=2sinx+1
（1）y=f(ωx)=2sinωx+1
要使y=f(ωx)在区间[-π/2,2π/3]上是增函数
则T≥4*(2π/3)=8π/3
T=2π/ω 解得：ω≤3/4
（2）f(x)的T为2π,A是B的子集
根据题意得,f(x)在π/6≤x≤2π/3之间的取值范围为[-1,3]
则 f(x)-m1
3、（1）f(x)=2cos^2 x+ 根号3 sin2x+α
=cos2x+1+ 根号3 sin2x+α
=2sin(x+π/6)+α+1 T=2π
则在[-π/6,π/6]上最大值与最小值之和分别在π/6和-π/6处取得
即2sin(π/6+π/6)+α+1+2sin(-π/6+π/6)+α+1=3
解得：α=(1-根号3）/2
（2）要使f(x)=0在[0,π]内有两相异实根x1 x2
则0

证明：
3-sin^4 x-cos^4 x
=1+(1-sin^4 x)+(1-cos^4 x)
=1+(1-sin^2 x)(1+sin^2 x)+(1-cos^2 x)(1+cos^2 x)
=1+(cos^2 x)(1+sin^2)+(sin^2 x)(1+cos^2 x)
=1+cos^2 x+sin^2 x+2(sin^2 x)(cos^2 x)
=2+2(sin^2 x)(cos^2 x)
所以
(3-sin^4 x-cos^4 x)/2cos^2 x
=[2+2(sin^2 x)(cos^2 x)]/2cos^2 x
=sec^2 x+sin^2 x
=1+tan^2 x +sin^2 x

f(x)=[2cos^3x+sin^2(2π-x)+sin(π/2+x)-3]/[2+2cos^2(π+x)+cos(-x)]
=(2cos^3x+sin^2x+cosx-3)/(2+2cos^2x+cosx)
=(2cos^3x+1-cos^2x+cosx-3)/(2+2cos^2x+cosx)
=(2cos^3x-cos^2x+cosx-2)/(2+2cos^2x+cosx)
=(cosx-1)(2x^2+cosx+2)/(2+2cos^2x+cosx)
=cosx-1