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

(1)因为f(x)=23sinxcosx+1?2sin2x=3sin2x+cos2x=2sin(2x+π6),故 函数f(x)的最小正周期为T=π. 由2kπ?π2≤2x+π6≤2kπ+π2,k∈Z,得f(x)的单调递增区间为[kπ?π3,kπ+π6],k∈Z.(2)根据条件得μ=2sin(4x+5π6),当x∈[0,π8]时,4x+5π6∈[56π,...

(1)f(x)=23sinxcosx-2sin2x+1=3sin2x+cos2x=2sin(2x+π6),∴T=2π2=π.∵x∈[0,π2],∴π6≤2x+π6≤7π6,∴当2x+π6=π2时,即x=π3时函数有最大值,最大值为2;当2x+π6=7π6时,即x=π2时,函数有最小值为-1;(2)f(x0)=65,x0∈[π4,π2],则sin(2x...

(I)因为f(x)=2sin2x+23sinxcosx+1=1?cos2x+3sin2x+1=2sin(2x?π6)+2由2kπ?π2≤2x?π6≤2kπ+π2(k∈Z)得kπ?π6≤x≤kπ+π3(k∈Z)所以f(x)的单调增区间是[kπ?π6,kπ+π3](k∈Z);(Ⅱ)因为0≤x≤π2,所以?π6≤2x?π6≤5π6所以?12≤sin(2x?π6)≤1所以f(x)=2sin(2...

∵f(x)=3sin2x+cos2x=2sin(2x+π6),∴f(x)=2sin(2x+π6),T=π故选B.

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)=3sin2x+cos2x=2sin(2x+π6)∴f(x)的最小正周期为T=2π2=π,令sin(2x+π6)=0,则x=kπ2?π12(k∈Z),∴f(x)的对称中心为(kπ2?π12,0),(k∈Z);(2)∵x∈[?π6,π3]∴?π6≤2x+π6≤5π6∴?12≤sin(2x+π6)≤1∴-1≤f(x)≤2∴当x=?π6时,f(x)的最...

(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.

(1)f(x)=cos2x+2 3 sinxcosx=cos2x+ 3 sin2x…(2分)=2( 1 2 cos2x+ 3 2 sin2x)=2sin(2x+ π 6 )…(4分)∴T=π,f(x) 最大值 =2…(6分)(2)先将y=2sinx(x∈R)的图象向左移 π 6 个单位,得到y=2sin(x+ π 6 )的图象;再将y=2sin(x+...

(Ⅰ)∵f(x)=23sinxcosx-2cos2x=3sin2x-(1+cos2x)=2sin(2x-π6)-1,∴函数f(x)的最小正周期T=π;由2kπ+π2≤2x-π6≤2kπ+3π2得:kπ+π3≤x≤kπ+5π6,k∈Z.∴函数f(x)的单调递减区间为[kπ+π3,kπ+5π6]k∈Z.(Ⅱ)∵x∈[0,π2],∴2x-π6∈[-π6,5π6],∴-...

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

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