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已知函数F(x)=23sinxCosx+2Cos2x%1(x∈R),g(x...

(Ⅰ) f(x)=23sinxcosx+2cos2x?1=3sin2x+cos2x=2sin(2x+π6)则g(x)=|2sin(2x+π6)|,∵y=|sinx|的单调递减区间为[kπ+π2,kπ+π],(k∈Z). ∴由kπ+π2≤2x+π6≤kπ+π 得:kπ2+π6≤x≤kπ2+5π12,则g(x)的单调递减区间为[kπ2+π6,kπ2+5π12](k∈Z). (...

(1)由f(x)=23sinxcosx+2cos2x-1,得f(x)=3(2sinxcosx)+(2cos2x-1)=3sin2x+cos2x=2sin(2x+π6),所以函数f(x)的最小正周期为π.(2)因为f(x)=2sin(2x+π6)在区间[0,π6]上为增函数,在区间[π6,π2]上为减函数,又f(0)=1,f(π...

f(x)=23sinxcosx+2cos2x?1,得f(x)=3(2sinxcosx)+(2cos2x?1)=3sin2x+cos2x=2sin(2x+π6)(1)T=2π2=π,所以函数f(x)的最小正周期为π;由2kπ?π2≤2x+π6≤2kπ+π2得kπ?π3≤x≤π6+kπ所以函数f(x)的单调递增区间为[kπ?π3,π6+kπ],k∈Z.(2)∵x∈...

(1)由数f(x)=23sinxcosx+2cos2x-1,得f(x)=3sin2x+cos2x=2sin(2x+π6),所以函数f(x)的最小正周期为π;∵2kπ-π2<2x+π6<2kπ+π2,k∈Z∴x∈(kπ-π3,kπ+π6),k∈Z又x∈[0,π2],f(x)=2sin(2x+π6)在[0,π2]上的单调递增区间为(0,π6);...

(1)f(x)=23sinx?cosx+2cos2x=3sin2x+cos2x+1=2(32sin2x+12cos2x)+1=2sin(2x+π6)+1,x∈R…(4分)∴f(x)的最小正周期为T=2π2=π.…(6分)(2)∵f(α2)=2sin[2(α2)+π6]+1=2sin(α+π6)+1=13,…(7分)∴sin(α+π6)=?13<0,…(8分)∵α∈[...

(Ⅰ)f(x)=3sin2x+cos2x=2sin(2x+π6)----(2分)∴T=π,-----(3分)当2kπ+π2≤2x+π6≤2kπ+3π2,k∈Z 时,即kπ+π6≤x≤kπ+2π3,k∈Z时,f(x)为减函数-----(5分)∴y=f(x)减区间为[kπ+π6,kπ+2π3],k∈Z-----(6分);(Ⅱ)当0≤x≤π2时,则π6≤2x+π6≤...

(1)因为f(x)=23sinxcosx+1?2sin2x=3sin2x+cos2x=2sin(2x+π6),故 函数f(x)的最小正周期为T=π. 由2kπ?π2≤2x+π6≤2kπ+π2,k∈Z,得f(x)的单调递增区间为[kπ?π3,kπ+π6],k∈Z.(2)根据条件得μ=2sin(4x+5π6),当x∈[0,π8]时,4x+5π6∈[56π,...

(1)由题知,f(x)=23sinxcosx+2cos2x-1=3sin2x+cos2x=2sin(2x+π6),∴函数的最小正周期为T=π;∵x∈[0,π2],∴2x+π6∈[π6,7π6],∴f(x)max=f(π6)=2 ,f(x)min=f(π2)=?1;(2)由(1)知,f(x0)=2sin(2x0+π6),∴f(x0)=2sin(2x0+π6)=65...

(I)∵f(x)=23sinxcosx+2sin2x-1=3sin2x-cos2x=2sin(2x-π6),∴函数f(x)的最小正周期为T=π;由-π2+2kπ≤2x-π6≤π2+2kπ,k∈Z,解得:-π6+kπ≤≤π3+kπ,k∈Z,则f(x)的单调递增区间为[-π6+kπ,π3+kπ],k∈Z;(II)函数y=f(x)的图象上各点的纵...

(1)函数f(x)=2cos2x+23sinxcosx?1(x∈R)=cos2x+3sin2x=2(12cos2x+32sin2x)=2sin(π6+2x),∴周期T=2πω=2π2=π.(2)由 2kπ-π2≤π6+2x≤2kπ+π2,k∈z,可得 kπ-π3≤x≤kπ+π6,故函数f(x)单调增区间为[kπ-π3,kπ+π6].(3)∵0≤x≤π2,∴π6≤π6+2x≤7π...

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