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已知函数F(x)=2Cosxsin(x+π3)?3sin2x+sinxCosx+2...

f(x)=2cosx(12sinx+32cosx)?3sin2x+sinxcosx=2sinxcosx+3(cos2x?sin2x)=sin2x+3cos2x=2sin(2x+π3)(1)则函数f(x)的最小正周期T=2π2=π;(2)由?π2+2kπ≤2x+π3≤π2+2kπ,解得?5π12+kπ≤x≤kπ+π12,k∈Z,即函数f(x)的单调递增区间[?5π12,kπ...

(1)由三角函数的公式化简可得f(x)=2cosx(12sinx+32cosx)-3sin2x+sinx?cosx=2sinxcosx+3cos2x?3sin2x=sin2x+3cos2x=2sin(2x+π3),由2kπ+π2≤2x+π3≤2kπ+3π2,可得kπ+π12≤x≤kπ+7π12,故f(x)的单调递减区间为[kπ+π12,kπ+7π12](k∈Z).(2...

(Ⅰ)f(x)=2cosx(12sinx+32cosx)?3sin2x+sinxcosx=2sinxcosx+3(cos2x?sin2x)=sin2x+3cos2x=2sin(2x+π3)∴T=π(Ⅱ)f(x)的减区间为2kπ+π2≤2x+π3≤2kπ+3π2,kπ+π12≤x≤kπ+7π12又∵x∈[?π2,?π12],∴?π2≤x≤?5π12或π12≤x≤π2即f(x)在[?π2,?5π12]和在[...

(1)f(x)=2cosxsin(x-π3)+3sin2x+sinxcosx=2cosx(12sinx-32cosx)+3sin2x+sinxcosx=sinxcosx-3cos2x+3sin2x+sinxcosx=2sinxcosx-3(cos2x-sin2x)=sin2x-3cos2x=2sin(2x-π3),∴T=2π2=π,(2)g(x)=f(

f(x)=2cosx(12sinx+32cosx)?3sin2x+sinx?cosx=2sinxcosx+3(cos2x?sin2x)…3′=sin2x+3cos2x=2sin(2x+π3)…5′(I)f(x)的值域为[-2,2]…(7分)(II)由题可知:g(x)=2sin(2(x?π6)+π3)=2sin2x,…(9分)∴2kπ?π2≤2x≤2kπ+π2,解得,kπ?π4≤x≤kπ...

(1)∵函数f(x)=2cosxsin(x+π3)?3sin2x+sinxcosx=sinxcosx+3cos2x-3sin2x+sinxcosx=sin2x+3cos2x=2sin(2x+π3).∵x∈[?π12,π6],∴π6≤2x+π3≤2π3,∴12≤sin(2x+π3)≤1,故函数f(x)的最大值为2,最小值为1.(2)锐角△ABC中,由f(A)=0 可得 s...

(I)令f(x)=0得sinx(3sinx+cosx)=0所以sinx=0或tanx=-33由sinx=0,x∈[π2,π]得x=π由tanx=-33,x∈[π2,π]得x=5π6综上所述,f(x)的零点为x=π或x=5π6(II)g(x)=f(x)?3sin2x=sinxcosx=12sin2x由2x=kπ+π2(k∈Z)得:x=kπ2+π4(k∈Z)即函数g...

f(x)=2cosxcos(x?π6)?3sin2x+sinxcosx=2cosx(32cosx+12sinx)-3sin2x+sinxcosx=3(cos2x-sin2x)+2sinxcosx=3cos2x+sin2x=2(32cos2x+12sin2x)=2sin(2x+π3),(1)∵ω=2,∴T=2π2=π;(2)∵f(x)=1,即2sin(2x+π3)=1,∴sin(2x+π3)=12,...

(1)函数f(x)=sinxcosx?3sin2x=12sin2x?3?1?cos2x2=12sin2x+32cos2x?32=sin(2x+π3)-32.则函数的周期T=2π2=π.(2)∵x∈[0,π4],∴2x+π3∈[π3,5π6],∴f(x)的最大值为f(π12)=1-32,最小值为f(π4)=12-32.

f(x)=2cosxcos(x一兀/6)一√3sin2x+sinxcosx =2cosx(√3/2cosx+1/2sinx)-√3sin2x+1/2sin2x =√3cos^2(x)+1/2sin2x-√3sin2x+1/2sin2x =√3cos^2(x) -√3sin2x+sin2x =√3/2(1+cos2x) -(√3-1)sin2x =√3/2+√3/2cos2x -(√3-1)sin2x 检查题目是否抄错?

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