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求函数y=2sin3x%1的最小正周期,最值,并求出y取得最...

所以该函数既没有最大值;2sinxlt,我是在word中解答的;2;0,或0lt;2sinx,则1,但是不等号和分数都无法粘贴过去,或3,k为整数};a;3/-3/2;2;2:如果alt,或y,且ab。这可怎么办啊?补上x后解答如下;1,取得极大值时的x的集合是{xx=3π/2所以该函数的...

f(x) = sin(π/2-x)sinx - √3cos²x = cosxsinx - √3cos²x = 1/2sin2x - √3/2cos2x - √3/2 = sin2xcosπ/3-cos2xsinπ/3 - √3/2 = sin(2x-π/3) - √3/2 最小正周期:2π/2 = π 最大值:1 - √3/2 = (2-√3)/2

f(x)=sin2x-2√3sin^2x+√3+1 =sin2x+√3(-2sin^2x+1)+1 =(sin2x+√3cos2x)+1 =(sin2xcos(π/3)+cos2xsin(π/3))*2+1 =2sin(2x+π/3)+1 最小正周期=π -π/2+2kπ

fx=1/2sin2x-√3cos²x =1/2sin2x+√3/2cos2x-√3/2 =sin2xcosπ/3+cos2xsinπ/3-√3/2 =sin(2x+π/3)-√3/2 最小正周期=2π/2=π 最小值=-1-√3/2

f(x)=√3sinxcosx-cos²x-1/2 =(√3/2)sin2x-(1+cos2x)/2-1/2 =(√3/2)sin2x-(1/2)cos2x-1 =sin(2x-π/6)-1. 早小正周期T=2π/2=π. sin(2x-π/6)=-1, 即x=kπ-π/6时, 最小值f(x)|min=-2。

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

(1)f(x)=3sin2x+cos2x=2(sin2xcosπ6+cos2xsinπ6)=2sin(2x+π6)∴T=2π2=π,当2x+π6=2kπ+π2,k∈Z,即x=π6+kπ,k∈Z时,函数取得最大值2.当2x+π6=2kπ-π2,即x=kπ-π3,k∈Z时,函数取得最小值-2.(2)当2kπ-π2≤2x+π6≤2kπ+π6,k∈Z时,即kπ-π3≤x...

(1)∵f(x)=5sin(k5x?π3)(k≠0)∴M=5,m=-5,T=2π|k5|=10π|k|;(2)由题意知,函数f(x)在任意两个整数间(包括整数本身)变化时,至少有一个值是M和一个值是m,∴T≤1,即10π|k|≤1,∴|k|≥10π>31.4,∵k∈N*,∴最小正整数k为32.

解答:(1)f(x)=2sin²(π/4+x)-根号3cos2x=1-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x+1=2sin(2x-π/3)+1∵x∈[π/4,π/2]∴2x-π/3∈[π/6,2π/3]∴sin(2x-π/3)∈[1/2,1]∴2x-π/3=π/6时,f(x)有最小值22x-π/3=π/2时,f(x)有最大值3(2)|f(x)-m|m-2且f(x)m-2...

(1)fx的最小正周期π (2)值域为【-1,1】 当2x-π/3=2kπ+π/2 x=kπ+5π/12,有最大值1 当2x-π/3=2kπ-π/2 x=kπ-π/12,有最小值-1

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