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已知F(x)=sin(2x+6/π)+2/3,x属于R

最小正周期2π/2=π 单调减区间 2kπ+π/2≤2x+π/6≤2kπ+3π/2 2kπ+π/3≤2x≤2kπ+4π/3 kπ+π/6≤x≤kπ+2π/3 y=sin(2x+π/6)+3/2 =sin[2(x+π/12)]+3/2 因此将sin2x向左移π/12得到sin[2(x+π/12)] 再向上移3/2个单位即可

(1)由函数 f(x)=sin(2x+ π 6 )+ 3 2 ,x∈R ,可得周期等于 T= 2π 2 =π.由 2kπ- π 2 ≤2x+ π 6 ≤2kπ+ π 2 (k∈Z) 求得 kπ- π 3 ≤x≤kπ+ π 6 (k∈Z) ,故函数的递增区间是 [ .(2)由条件可得 f(x)=sin(2x+ π 6 )+ 3 2 =sin[2(x+ π 12 )]+ 3 2 ....

f(x) =sin(2x+π/6)+3/2 2kπ+π/2≤2x+π/6≤2kπ+3π/2 k∈Z 2kπ+π/6≤x≤2kπ+2π/3 因此,f(x)的单减区间为:[2kπ+π/6,2kπ+2π/3] (k∈Z)

21.(3)x∈[0,π/3], u=2x+π/6的值域是[π/6,5π/6], v=f(x)=sinu-√3的值域是[1/2-√3,1-√3], 对任意x∈[0,π/3],f(x)^2-(2+m)f(x)+2+m

1. f(x)=sin(2x+π/6)+1/2 则最小正周期为T=2π/2=π。 2. y=(1/2)(cosx)^2+(√3/2)sinxcosx+1 =(√3/4)sin2x+(1/4)cos2x+5/4 =(1/2)[(√3/2)sin2x+(1/2)cos2x]+5/4 =(1/2)(sin2xcosπ/6+cos2xsinπ/6)+5/4 =(1/2)sin(2x+π/6)+5/4 f(x)最小正周期为T=2π/...

(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...

函数f(x)在区间[-π/2,-π/12]是连续且可导的。其导数f'(x)=6cos(2x+π/6), Then cos(2x+π/6)=0; 2x+π/6=π/2, -π/2, -3π/2, 在[-π/2,-π/12]区域中只有x1=-π/3 符合 f(x)在-π/3处有极小-3,-π/12处值最大,3sin(π/6)

解: f(x)=sin2xcosπ/6+cos2xsinπ/6+sin2xcosπ/6-cos2xsinπ/6+1+cos2x =2sin2xcosπ/6+cos2x+1 =√3sin2x+cos2x+1 =2sin(2x+π/6)+1 ⑴f(x)取得最大值3,此时2x+π/6=π/2+2kπ,即x=π/6+kπ,k∈Z 故x的取值集合为{x|x=π/6+kπ,k∈Z} ⑵由2x+π/6∈[-π/2+2kπ...

(Ⅰ)∵f(x)=3sin(2x-π6)+1-cos(2x-π6)=1+2sin(2x-π3),∵ω=2,∴函数f(x)的最小正周期为π;(Ⅱ)∵x∈[-π4,π4],∴2x-π3∈[-5π6,π6],∴-1≤sin(2x-π3)≤12,∴当x∈[-π4,π4]时,f(x)max=2,f(x)min=-1.

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