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已知F(x)=sin(2x+π6),x∈R.(1)求函数F(x)...

(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...

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(2x+π/6)+3/2 -1

(1)∵函数f(x)=sin(2x-π6),∴f(π4)=sinπ3=32.(2)当且仅当2x-π6=2kπ+π2,k∈z时,即x=kπ+π3时,该函数取得最大值1,所以该函数取得最大值时自变量的取值集合为{x|x=kπ+π3,k∈z}.(3)由f(α+π3)=35,求得cos2α=35=1-2sin2α,∴sinα=±55...

(1)∵f(x)=2sin(2x+ π 6 ),∴其最小正周期T= 2π 2 =π;∴由2kπ- π 2 ≤2x+ π 6 ≤2kπ+ π 2 得kπ- π 3 ≤x≤kπ+ π 6 (k∈Z),∴函数的增区间为[kπ- π 3 ,kπ+ π 6 ](k∈Z),(2)∵x∈( π 4 , 3π 4 ],∴2x+ π 6 ∈( 2π 3 , 5π 3 ],∴-1≤sin(2...

参考

(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 ....

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/...

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)=sin2x?cos2xcosπ6?sin2xsinπ6=12sin2x?32cos2x=sin2xcosπ3?cos2xsinπ3=sin(2x?π3)∴最小正周期T=2π2=π(2)由题意,解不等式?π2+2kπ≤2x?π3≤π2+2kπ得 ?π12+kπ≤x≤5π12+kπ,(k∈Z)∴f(x)的递增区间是[?π12+kπ,5π12+kπ](k∈Z)

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