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

(1)f(x)=sin2x+3sinxsin(x+π2)=1?cos2x2+32sin2x=sin(2x-π6)+12∴当π2+2kπ≤2x-π6≤3π2+2kπ(k∈Z),时函数单调减,即2π3+kπ≤x≤5π3+kπ(k∈Z),函数单调减,∴函数的单调递减区间为[2π3+kπ,5π3+kπ](k∈Z),(2)∵f(x)=sin(2x-π6)+12∴...

(Ⅰ) f(x)=3sinxcosx+sin2x=32sin2x-12cos2x+12=sin(2x-π6)+12,∵ω=2,∴T=2π2=π,则函数f(x)的最小正周期是π; (Ⅱ)∵x∈[0,π2],∴2x-π6∈[-π6,5π6],∴sin(2x-π6)∈[-12,1],即sin(2x-π6)+12∈[0,32],则f(x)在[0,π2]上的最大值和...

最小正周期2π/2=π 单调减区间 2kπ+π/2≤2x+π/6≤2kπ+3π/2 2kπ+π/3≤2x≤2kπ+4π/3 kπ+π/6≤x≤kπ+2π/3 y=sin(2x+π/6)+3/2 =sin[2(x+π/12)]+3/2 因此将sin2x向左移π/12得到sin[2(x+π/12)] 再向上移3/2个单位即可

(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...

(1)列表: 2x+π3 0 π2 π 3π2 2π x -π6 π12 π3 7π12 5π6 f(x) 0 2 0 -2 0画出函数的图象:(2)令 2kπ+π2≤2x+π3≤2kπ+3π2,k∈z,可得 kπ+π12≤2x+π3≤kπ+7π12,k∈z.故函数f(x)的单调递减区间为[kπ+π12,kπ+7π12],k∈z.

(1)∵f(x)=sin2x?cosπ3+cos2x?sinπ3+sin2x?cosπ3-cos2x?sinπ3+cos2x=sin2x+cos2x=2sin(2x+π4),∴函数f(x)的最小正周期T=2π2=π.(2)∵函数f(x)在区间[?π4,π8]上是增函数,在区间[π8,π4]上是减函数,又f(-π4)=-1,f(π8)=2,f(π4...

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

(Ⅰ)函数f(x)=sin2x+2sinxsin(π2?x) +3sin2(3π2?x)=sin2x+2sinxcosx+3cos2x=1+sin2x+2cos2x=2+sin2x+cos2x=2+2sin(2x+π4)∵tan2x=43,所以sin2x=45,cos2x=35或sin2x=?45,cos2x=?35∴f(x)=175或35(Ⅱ)∵x∈[0,π2],∴2x+π4∈[π4,3π4],2+...

(1)f(x)=2cosxsin(x+π3)?3sin2x+sinxcosx+2=2sin(2x+π3)+2∴最小正周期T=2π2=π,当2kπ-π2≤2x+π3≤2kπ+π2时,即kπ-5π12≤x≤kπ+π12,函数单调增∴函数的单调增区间为:[kπ-5π12,kπ+π12](k∈Z)(2)由函数y=sinx纵坐标不变,横坐标扩大2倍得到y=...

(1)f(x)=3(1?cos2x)2+12sin2x?32=12sin2x?32cos2x=sin(2x-π3)∵x∈(0,π2)∴?π3<2x?π3<2π3.∴当?2x?π3=π2时,即x=5π12时,f(x)的最大值为1.(2)由f(x)=sin(2x-π3),若x是三角形的内角,则0<x<π,∴?π3<2x?π3<5π3.令f(x)=12...

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