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已知函数F(x)=sin2x?Cos(2x?π6),其中x∈R.(1)...

(I)∵函数f(x)=sin(7π6?2x)+2cos2x?1=sin7π6cos2x-cos7π6sin2x+cos2x=32sin2x+12cos2x=sin(2x+π6).故函数f(x)的周期为T=2π2=π.再令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,求得 kπ-π3≤x≤kπ+π6,k∈z,故单调递增区间为[kπ-π3,kπ+π6],k∈z.(II)...

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...

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...

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)=sin(2x-π6)+2cos2x-1=32sin2x-12cos2x+cos2x=32sin2x+12cos2x=sin(2x+π6)由-π2+2kπ≤2x+π6≤π2+2kπ(k∈Z)得:-π3+kπ≤x≤π6+kπ(k∈Z)由π2+2kπ≤2x+π6≤3π2+2kπ(k∈Z)得:π6+kπ≤x≤2π3+kπ(k∈Z)故f(x)的单调递增区间是[-π3+kπ,π6+kπ]...

稍等

(1)f(x)=1+cos2x+sin2x+a=2sin(2x+π4)+1+a,∵ω=2,∴T=π,∴f(x)的最小正周期π;当2kπ-π2≤2x+π4≤2kπ+π2(k∈Z)时f(x)单调递增,解得:kπ-3π8≤x≤kπ+π8(k∈Z),则x∈[kπ-3π8,kπ+π8](k∈Z)为f(x)的单调递增区间;(2)当x∈[0,π6]时,π...

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

(1)f(x)=cosx(asinx-cosx)+cos2(π2+x)=a2sin2x-cos2x,由f(?π3)=f(0),解得a=23故f(x)=a2sin2x-cos2x=3sin2x-cos2x=2sin(2x-π6),故单调递减区间为[kπ+π3,kπ+5π6],k∈Z(2)由a2+c2?b2a2+b2?c2=c2a?c,可解得2sinAcosB=sinA,∵s...

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