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已知F(x)=2sin(1/2x%π/6)%3,x∈R求①函数F(x)的最小...

(1)T=2π2=π.(2)由2kπ-π2≤2x+π6≤2kπ+π2,得kπ-π3≤x≤kπ+π6,k∈Z,∴函数的单调增区间为[kπ-π3,kπ+π6](k∈Z).(3)∵x∈[0,π2],∴2x+π6∈[π6,7π6],∴-12≤sin(2x+π6)≤1,∴当2x+π6=π2,即x=π6时函数有最大值1,当2x+π6=7π6时,即x=π2,函数有...

见图 解:(I)f(x)==sin2x+cos2x=sin(2x+). 令 2kπ-≤(2x+)≤2kπ+,可得 kπ-≤x≤kπ+,k∈z. 即f(x)的单调递增区间为[kπ-,kπ+],k∈z. (II)在△ABC中,由,可得sin(2A+)=,∵<2A+<2π+, ∴<2A+= 或,∴A= (或A=0 舍去). ∵b,a,c成...

(1)∵f(x)=sin(2x+π6)+32,∴当2x+π6=2kπ-π2,k∈z,即x=kπ-π3时,函数f(x)取得最小值为-1+32=12.(2)令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,求得 kπ-π3≤x≤kπ+π6,故函数的增区间为[kπ-π3,kπ+π6],k∈z.(3)把函数y=sinx的图象向左平移π6个单位...

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

f(x)=sin(7π6?2x)?2sin2x+1=?12cos2x+32sin2x+cos2x=12cos2x+32sin2x=sin(2x+π6)…(3分)(1)最小正周期:T=2π2=π,…(4分)由2kπ?π2≤2x+π6≤2kπ+π2(k∈Z)可解得:kπ?π3≤x≤kπ+π6(k∈Z),所以f(x)的单调递增区间为:[kπ?π3,kπ+π6](k∈Z); …...

解:先用降幂公式把函数化为:f(x)=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1 (1)最小值为-2,最小正周期为π (2)由f(C)=0知sin(2C-π/6)=1,从而可得C=π/3,再由余弦定理知:c^2=a^2+b^2-2abcosC 3=a^2+4a^2-2a*2acosπ/3,解得a=1,故b=2

f(x)=1/2sin(2x+6π)+5/4. 当f(x)单调递增时, 2kπ-π/2≤2x+6π≤2kπ+π/2 →kπ-13π/4≤x≤kπ-11π/4. 即单调递增区间为: [kπ-13π/4,kπ-11π/4]. 2x+6π=kπ+π/2 →x=kπ/2-11π/4. 故对称轴为: x=kπ/2-11π/4, 对称中心为: (kπ/2-11π/4,0). 当f(x)取最小值时...

解如图。

(1) y=1-(1/2)cos[(π/3)x],x属于R y(-x)=1-(1/2)cos[(π/3)(-x)]=1-(1/2)cos[(π/3)x]=y(x),故是 . T=2π/(π/3)=6; 单增区间:由-π+2kπ≦(π/3)x≦2kπ,得单增区间为 -3+6k≦x≦6;K∈Z 单减区间:由2kπ≦(π/3)x≦π+2kπ,得单减区间为 6k≦x≦3+6k;k∈Z 最大值...

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