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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...

(Ⅰ)f(x)=3cos2x+2sin(3π2+x)sin(π-x)=3cos2x-2cosxsinx=3cos2x-sin2x=2(32cos2x-12sin2x)=2cos(2x+π6),∴T=2π2=π,令2x+π6=kπ(k∈Z),即x=kπ2-π12(k∈Z),∴函数f(x)的对称轴方程为x=kπ2-π12(k∈Z),(Ⅱ)∵f(x)=2cos(2x+π6)...

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)=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...

(1)∵f(x)=[2sin(x+π3)+sinx]cosx?3sin2x =2sinxcosx+3cos2x?3sin2x =sin2x+3cos2x=2sin(2x+π3).∴最小正周期T=2π2=π.(2)∵x 0∈[0,5π12],∴2x 0+π3∈[23,7π6],∴sin(2x 0+π3)∈[?12,1],∴f(x0)的值域为[-1,2].∵存在x 0∈[0,5π12],使...

(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...

(1)f(x)=sin2x+2sinxcosx+3cos2x=sin2x+cos2x+2=2sin(2x+π4)列表: x ?π2 ?3π8 ?π8 π8 3π8 π2 2x+π4 ?3π4 ?π2 0 π2 π 3π4 f(x) -1 -2 0 2 0 -1描点、连线:(2)函数y=2sin2x的图象向左平移π8个单位得到函数y=2sin2(x+π8)的图象,即得到...

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