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已知函数F(x)=3sin2x+sinxCosx?32(x∈R).(1)若x...

(1)∵sin25π6=12,cos25π6=32,∴f(25π6)=-3sin225π6+sin25π6cos25π6=0.(2)f(x)=32cos2x-32+12sin2x.∴f(α2)=32cosα+12sinα-32=14-3216sin2α-4sinα-11=0,解得sinα=1±358∵α∈(0,π),∴sinα>0故sinα=1+358.

f(x)=sin²x+sinxcosx+1 =(1-cos2x)/2+sin2x/2+1 =½(sin2x-cos2x)+3/2 =√2/2sin(2x-π/4)+3/2 ∴最小正周期是π 单调递增区间2x-π/4∈(2kπ-π/2,2kπ+π/2)→x∈(kπ-π/8,kπ+3π/8) 单调递减区间2x-π/4∈(2kπ+π/2,2kπ+3π/2)→x∈(kπ+3π/8,kπ+7π/8)

(法一)f(x)=3×1+c0s2x2+12sin2x?32=12sin2x+32cos2x=sin(2x+π3)(I )f(π4)=sin(2×π4+π3 )=cosπ3=12(II)函数的基本性质如下①奇偶性:函数f(x)既不是奇函数,也不是偶函数②单调性:函数f(x)的单调增区间为[kπ?5π12,kπ+π12] k∈Z,单...

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

已知函数f(x)=√3sin^2x+sinxcosx+m,x∈[0,兀/2] √3sin^2x+sinxcosx =√3(1+cos2x)/2+(sin2x)/2 =√3/2+√3(cos2x)/2+(sin2x)/2 =√3/2+cos(2x-兀/6) x∈[0,兀/2] 当x=兀/12时有最大值为√3/2+1,当x=兀/2时有最小值为0 则函数f(x)=√3sin^2x+sinxco...

解(1)∵f(x)=2sin2x+23sinxcosx-1=3sin2x-cos2x=2(sin2xcosπ6-cos2xsinπ6),∴f(x)=2sin(2x-π6).∴函数f(x)的图象可由y=sinx的图象按如下方式变换得到:①将函数的y=sinx图象向右平移π6个单位,得到函数y=sin(x-π6)的图象;②将函数y=...

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

f(x)=2cosxcos(x一兀/6)一√3sin2x+sinxcosx =2cosx(√3/2cosx+1/2sinx)-√3sin2x+1/2sin2x =√3cos^2(x)+1/2sin2x-√3sin2x+1/2sin2x =√3cos^2(x) -√3sin2x+sin2x =√3/2(1+cos2x) -(√3-1)sin2x =√3/2+√3/2cos2x -(√3-1)sin2x 检查题目是否抄错?

F(x)=sin²x+√3sinxcosx+2cos²x =(√3/2)sin2x+1+(1+cos2x)/2 =(√3/2)sin2x+(1/2)cos2x+3/2 =sin(2x+π/6)+3/2 所以最小正周期是T=2π/2=π 令2kπ-π/2<2x+π/6<2kπ+π/2,k∈Z 2kπ-2π/3<2x<2kπ+π/3,k∈Z kπ-π/3<x<kπ+π/6,k∈Z 所以单调递增...

∵sinx+cosx=2sin(x+π4)≠0,∴x+π4≠kπ即x≠kπ-π4,故①错误;∵f(x)=1+sin2xsinx+cosx=|sinx+cosx|sinx+cosx=±1,∴f(x)的值域为{-1,1},故②错误;∵f(x+2π)=1+sin2(x+2π)sin(x+2π)+cos(x+2π)=1+sin2xsinx+cosx=f(x),∴f(x)是周期函数,又f...

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