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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

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

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】

(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...

f(x)=2√3sinxcosx+2sin^2x-1=√3sin2x-cos2x=2sin(2x-π/6) 最小正周期T=π ,单调递增区间:2kπ-π/2

(1) f(x)=sin^2x+√3sinxcosx+1/2 =-1/2(1-2sin^2x)+√3/2sin2x+1 =cos(π/6)sin2x- sin(π/6)cos2x+1 = sin(2x-π/6)+1 f(x)的最小正周期=2π/2=π (2) f(x)= sin(2x-π/6)+1 2x-π/6=π/2+2kπ x=π/3+kπ f(x)=2为最大值 2x-π/6=-π/2+2kπ x=-π/6+kπ f(x)=0...

y=3(1-cosx)/2+1 =5/2-3cosx/2 所以cosx=-1 最大是4 x系数是1 T=2π 选A

f(x)=(√3/2)sinωx-(1/2)sinωx+1/2 =sin(ωx-π/6)+1/2. T=2π/ω=π,则ω=2. f(x)=sin(2x-π/6)-1/2单调增,则 2kπ-π/2≤2x-π/6≤2kπ+π/2, 即kπ-π/6≤x≤kπ+π/3, 即单调递增区间为[kπ-π/6,kπ+π/3]; f(x)=sin(2x-π/6)-1/2单调递减,则 2kπ+π/2≤2x-π/6≤2k...

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