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已知函数F(x)=sin2x+3sinxsin(x+π2),x∈R.(1...

(Ⅰ)∵f(x)=3sinxcosx-sin2x+12=32sin2x+12cos2x=sin(2x+π6),∴函数f(x)的最小正周期T=π…(4分)当2x+π6=2kπ+π2(k∈Z),即x=kπ+π6(k∈Z)时,f(x)max=1,∴当f(x)取得最大值时自变量x的集合为{x|x=kπ+π6,k∈Z};…(2分)(Ⅱ)g(x)=f...

f(x)=2sin²x+2√3sinx×sin(x+π/2) =1-cos2x+2√3sinxcosx =1-cos2x+√3sin2x =2(√3/2*sin2x-1/2*cos2x)+1 =2sin(2x-π/6)+1 最小正周期:T=2π/2=π 行家正解,不明白可以追问!祝您学习进步 满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢

1、 f(x)=sinx(sinxcosπ/6+cosxsinπ/6) =√3/2*sin²x+1/2*sinxcosx =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4+1/2*(1/2*sin2x-√3/2*cos2x) =√3/4+(sin2xcosπ/3-cos2xsinπ/3) =√3/4+sin(2x-π/3) 所以T=2π/2=π 2、 0≤x≤π/2 -π/3≤2x-π/3≤2π/3 -√3...

(I)f(x)=1?cos2x2+32sin2x+(1+cos2x)=32sin2x+12cos2x+32=sin(2x+π6)+32.∴f(x)的最小正周期T=2π2=π.由题意得2kπ?π2≤2x+π6≤2kπ+π2,k∈Z,即kπ?π3≤x≤kπ+π6,k∈Z.∴f(x)的单调增区间为[kπ?π3,kπ+π6],k∈Z.(II)先把y=sin2x图象上所有...

因为f(x)=[2sin(x+π3)+sinx]cosx?3sin2x=[2(sinxcosπ3+sinπ3cosx)+sinx]cosx-3sin2x=sin2x+3cos2x=2sin(2x+π3)于是(I)函数f(x)的最小正周期T=2π2=π.(II)因为x∈[0,π4]∴π3≤2x+π3≤5π6∴12≤sin(2x+π3)≤ 1即:1≤y≤2∴f(x)max=2,f(x)min=1

由五点法,列表:描点画图,如下:这种曲线也可由图象变换得到,即:y=sinx左移π3个单位y=sin(x+π3)纵坐标不变,横坐标变为12倍y=sin(2x+π3)纵坐标变为3倍,横坐标不变y=3sin(2x+π3)

(1)解法一:∵f(x)=1?cos2x2+sin2x+3(1+cos2x)2=2+sin2x+cos2x=2+2sin(2x+π4)(4分)∴当2x+π4=2kπ+π2,即x=kπ+π8(k∈Z)时,f(x)取得最大值2+2.因此,f(x)取得最大值的自变量x的集合是{x|x=kπ+π8,k∈Z}. (8分)解法二:∵f(x)=(sin2...

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...

f(x)=cos(x)-2sin2x+1,当x=0时,最大,f(x)=1-0+1=2;第二问找出x0/2和cos2x0的关系即可。希望可以帮到你

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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