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已知函数F(x)=23sinxCosx+2sin2x?1,x∈R.(I)求...

(Ⅰ)∵f(x)=23sinxcosx+2sin2x+3=2sin(2x-π6)+4,∴函数f(x)的最小正周期T=π,由2kπ-π2≤2x-π6≤2kπ+π2,即kπ-π6≤x≤kπ+π3,即函数的单调递增区间是[kπ-π6,kπ+π3];(Ⅱ)若f(α2)=265,得sin(α-π6)=35;sin(2α+π6)=cos[(2α+π6)-π2]=co...

解(1)∵f(x)=2sin2x+23sinxcosx-1=3sin2x-cos2x=2(sin2xcosπ6-cos2xsinπ6),∴f(x)=2sin(2x-π6).∴函数f(x)的图象可由y=sinx的图象按如下方式变换得到:①将函数的y=sinx图象向右平移π6个单位,得到函数y=sin(x-π6)的图象;②将函数y=...

(1)f(x)=cos2x?sin2x+23sinxcosx+1=3sin2x+cos2x+1=2sin(2x+π6)+1.(4分)因此f(x)的最小正周期为π,最小值为-1.(6分)(2)由F(a)=2得2sin(2α+π6)+1=2,即2sin(2x+π6)=12,而由a∈[π4,π2],得2a+π6∈[23π,76π].(9分)故2a+π6=56...

(I)f(x)=sin2x-3cos2x+1=2sin(2x-π3)+1,∵ω=2,∴T=π;(II)∵x∈[π4,π2],2x-π3∈[π6,2π3],∴sin(2x-π3)∈[12,1],即2sin(2x-π3)+1∈[2,3],则函数f(x)的最大值为3,最小值为2.

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

f(x)=sin²x+2sinxcosx+3cos²x =(1-cos2x)/2+sin2x+3(1+cos2x)/2 =√2sin(2x+π/4)+2 (I) 函数f(x) 的最小正周期T=2π/2=π ∵x∈(0,π) ∴2x+π/4∈(π/4,9π/4) 当2x+π/4=π/2.即x=π/x时,f(x)=2+√2为最大值, 当2x+π/4=3π/2.即x=5π/8时,f(x)...

f(x)=√3sinxcosx-cos²x-1/2 =(√3/2)sin2x-(1+cos2x)/2-1/2 =(√3/2)sin2x-(1/2)cos2x-1 =sin(2x-π/6)-1. 早小正周期T=2π/2=π. sin(2x-π/6)=-1, 即x=kπ-π/6时, 最小值f(x)|min=-2。

(法一)f(x)=3×1+c0s2x2+12sin2x?32=12sin2x+32cos2x=sin(2x+π3)(I )f(π4)=sin(2×π4+π3 )=cosπ3=12(II)函数的基本性质如下①奇偶性:函数f(x)既不是奇函数,也不是偶函数②单调性:函数f(x)的单调增区间为[kπ?5π12,kπ+π12] k∈Z,单...

f(x)=2√3sinxcosx-2sin²x+1 =√3sin2x+cos2x =2sin(2x+π/6) (2) 2kπ-π/2

F(x)=sin²x+√3sinxcosx+2cos²x =(√3/2)sin2x+1+(1+cos2x)/2 =(√3/2)sin2x+(1/2)cos2x+3/2 =sin(2x+π/6)+3/2 所以最小正周期是T=2π/2=π 令2kπ-π/2<2x+π/6<2kπ+π/2,k∈Z 2kπ-2π/3<2x<2kπ+π/3,k∈Z kπ-π/3<x<kπ+π/6,k∈Z 所以单调递增...

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