求函数y=2 1 2cosx的最大值.最小值,并求取得最值时x的集合

由题知,y*(cosx+3)=cosx-3则：cosx=-3(y+1)/(y-1)由cosx的取值范围知：-1≤cosx≤1所以-1≤-3(y+1)/(y-1)≤1由-3(y+1)/(y-1)≥-1解

y=√5(1/√5sinx+2/√5cosx)+2=√5(sinx+α)+2所以函数y=sinx+2cosx+2的值域是[2-√5,2+√5]

y=cosx(cosx+sinx)=cos²x+sinxcosx=(cos2x+1)/2+1/2·sin2x=1/2·(sin2x+cos2x)+1/2=1/2·√2(√2/2·sin2x+

cosx不等于0x不等于kπ（k为整数）再问：能不能一步一步来再答：因为cosx为分母所以cosx不等于0cosx等于0时，········x=-360度x=-180度x=-0度x=180度x=-36

cosx-1≥0x=2kπk∈Z

y'=((x^2)'cosx-x^2(cosx)')/cosx^2=(2xcosx+x^2sinx)/cosx^2

y=(1-cosx)/sinxy'=[(1-cosx)'sinx-(1-cosx)(sinx)']/sin^2x=(sin^2x+cos^2x-cosx)/sin^2x=(1-cosx)/sin^2x

只需sinx－cosx≠0,即tanx≠1,定义域是{x|x≠kπ＋π/4},其中k是整数.补充：sinx－cosx=√2[sinxcos(π/4)－cosxsin(π/4)]=√2sin(x－π/4

y=－（cosX-√3/2）^2+2最小值（1－4√3）/4x=2k兀+兀最大值2x=2k兀+兀/6

原式=sinx+cosx+2sinxcosx+2cosx=1+sin2x+cos2x+1=2+根号2sin(2x+π/4)剩下的你应该会了求采纳

解y=1+cosx-sinx=√2(√2/2cosx-√2/2sinx)+1=√2(cosxcosπ/4-sinxsinπ/4)+1=√2cos(x+π/4)+1∴周期为：T=2π/1=2π∵cos(

y'=lim(Δx→0)(cos(x+Δx)-cosx)/Δx=lim(Δx→0)(cosxcosΔx-sinxsinΔx-cosx)/Δx=lim(Δx→0)(cosxcosΔx-cosx)/Δx-

y=sinxcosx-2(sinx+cosx)+4=[(sinx+cosx)^2-1]/2-2(sinx+cosx)+4令sinx+cosx=t,-√2≤t≤√2y=t^2/2-2t+7/2=(t-2

y=cos^2x+sinx=1-sin^2x+sinx=-(sin^2x-sinx+1/4)+1+1/4=-(sinx-1/2)^2+5/4当sinx=1/2时有最大值y=5/4

sinx-cosx=√2(√2/2*sinx-√2/2cosx)=√2(sinxcosπ/4-cosxsinπ/4)=√2sin(x-π/4)

y=sinx^2+2sinx*cosx+3cosx^2=sinx^2+cosx^2+2sinx*cosx+2cosx^2=1+2cosx^2+sin2x=sin2x+cos2x=√2sin（2x+π/

y={cosx+cosx=2cosx;x∈[-π/2+2kπ,π/2+2kπ]-cosx+cosx=0;x∈(π/2+2kπ,π/2+2kπ)k∈Z.利用三角函数与0的关系分析.

第一个的定义域为一切实数,第二个只要tanx不等于0,即是x不等于kπ,k为0或者整数

y=(根号2)*sin[x+(π/4)]当[x+(π/4)]=(π/2)+2kπ,解得x=(π/4)+2kπ.k=0,1,……时,y达到最大(根号2)当[x+(π/4)]=-(π/2)+2kπ,解得x