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

(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)=23sinx?cosx+2cos2x=3sin2x+cos2x+1=2(32sin2x+12cos2x)+1=2sin(2x+π6)+1,x∈R…(4分)∴f(x)的最小正周期为T=2π2=π.…(6分)(2)∵f(α2)=2sin[2(α2)+π6]+1=2sin(α+π6)+1=13,…(7分)∴sin(α+π6)=?13<0,…(8分)∵α∈[...

(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...

(1)∵函数f(x)=23sinxcosx?2sin2x+1=3sin2x+cos2x=2sin(2x+π6),令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,解得 kπ-π3≤x≤kπ+π6,k∈z.故函数f(x)的单调增区间为[kπ-π3,kπ+π6],k∈z.(2)∵x∈[0,π2],∴2x+π6∈[π6 ,7π6],故当2x+π6=7π6,即x=π2时,...

(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+...

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

(1)由数f(x)=23sinxcosx+2cos2x-1,得f(x)=3sin2x+cos2x=2sin(2x+π6),所以函数f(x)的最小正周期为π;∵2kπ-π2<2x+π6<2kπ+π2,k∈Z∴x∈(kπ-π3,kπ+π6),k∈Z又x∈[0,π2],f(x)=2sin(2x+π6)在[0,π2]上的单调递增区间为(0,π6);...

(1)∵f(x)=sin2x+23sinxcosx+3cos2x,∴f(x)=1+3sin2x+2cos2x=3sin2x+cos2x+2=2sin(2x+π6)+2,故函数的最小正周期为T=2π2=π,由2kπ-π2≤2x+π6≤2kπ+π2,可得kπ-π3≤x≤kπ+π6,故函数单调递增区间为:[kπ-π3,kπ+π6](k∈Z)(2)∵x∈[-π6,π3],...

(1)由f(α)=5,得3sin2α+23sinαcosα+5cos2α=5.∴31?cos2α2+3sin2α+51+cos2α2=5.∴3sin2α+cos2α=1,即3sin2α=1?cos2α?23sinαcosα=2sin2αsinα=0或tanα=3,∴tanα=0或tanα=3.(5分)(2)由2accosB2abcosC=c2a?c,即cosBbcosC=12a?c,得

(1)f(x)=cos2x+3sin2x=2(sinπ6cos2x+cosπ6sin2x)=2sin(2x+π6)∴M=2,T=2π2=π(2)当2x+π6∈[π2+2kπ,3π2+2kπ],即x∈[π6+kπ,2π3+kπ](k∈R)时,f(x)单调递减.∴f(x)的单调递减区间为[π6+kπ,2π3+kπ](k∈R).

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