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y=2sin3x+1的最小正周期是?

sinx, 最小正周期为2π; 同理,sin3x,最小正周期为:3x=2π,x=2π/3 望采纳

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...

y = 3sin²(x/2) + 1 = - (3/2)cosx + 5/2 max(y) = 4 min(y) = 1 T = 2π

y=sin3x,最小正周期T=2π/|3|=(2/3)π, y=cos2x,最小正周期T=2π/|2|=π, 两者最小公倍数(除以这两个数的商都是正整数的最小正数)是2π, 所以y=sin3x+cos2x的最小正周期T=2π。

(1)∵f(x)=2sin(2x+π3).∴f(x)的最小正周期T=2π2=π;(2)用五点作图法作出f(x)的简图.列表: 2x+π3 0 π2 π 3π2 2π x ?π6 π12 π3 7π12 5π6 2sin(2x+π3) 0 2 0 -2 0函数的在区间[?π6,5π6]上的图象如下图所示:

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

原题是:已知函数y=2sinxcosx+2√3cos²x-√3. 1)求函数y的最小正周期和单调区间; 2)…… (1)y=sin2x+(√3)(1+cos2x)-√3 =sin2x+(√3)cos2x =2sin(2x+π/3) 即y=2sin(2x+π/3) T=(2π)/2=π 单增:2kπ-π/2≤2x+π/3≤2kπ+π/2,k∈Z kπ-5π/12≤x≤kπ+π/12...

(1)y=sin2x+2sinxcosx+3cos2x=(sin2x+cos2x)+sin2x+2cos2x=1+sin2x+(1+cos2x)=sin2x+cos2x+2=2sin(2x+π4)+2,∴函数的最小正周期T=2π2=π.(2)由2kπ?π2≤2x+π4≤2kπ+π2,得kπ?3π8≤x≤kπ+π8(k∈Z),∴函数的增区间为[kπ?3π8,kπ+π8](k∈Z).

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) y = sin(x + π/3) 最小正周期:T = 2π 值域:y ∈ [-1 , 1] 单调增区间:2kπ - π/2 ≤ x + π/3 ≤ 2kπ + π/2 , x ∈[2kπ - 5π/6 , 2kπ + π/6] 单调减区间:2kπ + π/2 ≤ x + π/3 ≤ 2kπ + 3π/2 , x ∈[2kπ + π/6 , 2kπ + 7π/6] (2) y = 2sin(1/2x -...

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