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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

y=a-bcos3x(b>0) -1≤-cosx≤1 -b≤-bcosx≤b ymax=a+b=3/2 ymin=a-b=-1/2 a=1/2,b=1 y=-4sin3bx=-4sin3x 3x属于(2kπ-π/2,2kπ+π/2)时单调减;3x属于(2kπ+π/2,2kπ+3π/2)时单调增 ∴单调减区间:(2kπ/3-π/6,2kπ/3+π/6) 单调增区间:(2kπ/...

(Ⅰ)∵{an}是首项为a1=14,公比q=14的等比数列,∴an=(14)n,bn+2=3log14an(n∈N*)=3log14(14)n=3n,∴bn=3n-2.(Ⅱ)cn=3bn?bn+1=3(3n?2)(3n+1)=13n?2?13n+1,∴Tn=(1?14)+(14?17)+…+(13n?2?13n+1)=1?13n+1.

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)=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

f(x)=2sinxcos(π/2-x)-√3sin(π+x)cosx+sin(π/2+x)cosx =2sinxsinx+√3sinxcosx+cosxcosx =2sin^2x+√3/2*sin2x+cos^2x =sin^2x+√3/2*sin2x+1 =(1-cos2x)/2+√3/2*sin2x+1 =√3/2*sin2x-1/2cos2x+1/2+1 =√3/2*sin2x-1/2cos2x+3/2 =sin2xcosπ/6-cos2xs...

(1)函数y=3sin(2x+π4) 的周期为2π2=π,令2kπ-π2≤2x+π4≤2kπ+π2,k∈z,求得 kπ-3π8≤x≤kπ+π8,故函数的增区间为[kπ-3π8,kπ+π8],k∈z.(2)当2x+π4=2kπ-π2,k∈z,即 x∈{x|x=kπ-3π8,k∈z}时,函数y取得最小值为-3.

T=2π÷2分子1=4π 最小值=-2 递增区间: 2kπ-π/2≤2分之x+3分之π≤2kπ+π/2 2kπ-5π/6≤2分之x≤2kπ+π/6 4kπ-5π/3≤x≤4kπ+π/3 即递增区间为【4kπ-5π/3,4kπ+π/3】

当sin3x=1,即自变量x的集合为 {x|3x=2kπ+ π 2 ,k∈z}={x|x= 2kπ 3 + π 6 ,k∈z} 时,函数y取得最大值为3.

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