kπ-2π/3

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 01:28:05
kπ-2π/3

kπ-2π/3
kπ-2π/3

kπ-2π/3
他这一步错了
(π/6-2X)属于(kπ-π/2,kπ+π/2)
应该是(π/6-2X)属于(2kπ-π/2,2kπ+π/2)
所以应该是
2kπ-2π/3<-2X<2kπ+π/3
两边同时除以-2
-kπ-π/6

做不来

你说的没错,是要除2的,不过那个人的答案是对的,过程中有问题,开始的时候应该是2kpi,除了2就是kpi了....

kπ-2π/3 下列终边相同的角是【选择题.】A.kπ+π/2与k*90°,k∈ZB.(2k+1)π与(4k±1)π,k∈ZC.kπ+π/6与2kπ±π/6,k∈ZD.(kπ)/3与kπ+(π/3),k∈Z 2kπ+π 与300°终边相同的是 A.kπ+π5/3(k∈z) B.2kπ-1π/3(k∈z) C.kπ与300°终边相同的是A.kπ+π5/3(k∈z)B.2kπ-1π/3(k∈z)C.kπ+6π/11(k∈z)D.2kπ+1π/3(k∈z) 函数y=3cos((π/3)-2x)的递减区间是A.[kπ-(π/2),kπ+(5π/12)] (k∈z)B.[kπ+(5π/12),kπ+(11π/12)](k∈z)C.[kπ-(π/3),kπ+(π/6)](k∈z)D.[kπ+(π/6),kπ+(2π/3)](k∈z) 不等式tanx≦-1的解集是选项:A.(2kπ-π/2,2kπ-π/4](k∈Z) B.[2kπ-π/4,2kπ+3π/2](k∈Z)C.(kπ-π/2,kπ-π/4](k∈Z) D.[2kπ+π/2,2kπ+3π/4](k∈Z) 若|cosa|=-cosa,则x取值范围A.2kπ≤x≤2kπ+π/2(k∈Z)B.2kπ+π/2≤x≤2kπ+3π/2(k∈Z)C.2kπ+3π/2≤x≤2kπ+2π(k∈Z)D.2kπ+π≤x≤2kπ+3π/2(k∈Z)最好有原因 函数y=sin(x-π/3)的单调递减区间是什么?A.【kπ-π/6,kπ+5π/6】(k属于Z)B.【2kπ-π/6,2kπ+5π/6】(k属于Z)C.【kπ-7π/6,kπ-π/6】(k属于Z)D.【2kπ-7π/6,2kπ-π/6】(k属于Z) 函数f(x)=sinx(sinx-cosx)的单调递减区间?A[2kπ+π/8,2kπ+5/8π] B[kπ+π/8,kπ+5/8π] C[2kπ-3/8π,2kπ+π/8] D[kπ-3/8π,kπ+π/8] (全部k属于Z) 6cos(2kπ+π/3)-2sin(2kπ+π/6)+3tan(2kπ) k€Z 第四象限角可以表示为1.(2kπ+3π/2,2kπ)k属于Z 2.(3π/2,2π)3.(2kπ—π/2,2kπ)k属于Z 4.(2kπ-3π/2,2kπ)k属于Z为什么第一个不对啊 若方程x^2 sina+y^2 sin2a =1,表示焦点在x轴上的椭圆,则a的取值范围为( )A.(kπ,kπ+π/2) ,k∈Z B.(2kπ,2kπ+π/2),k∈Z C.(2kπ,2kπ+π/3),k∈ZD.以上皆不正确 怎么证明tan(π/(2k+1))×tan(2π/(2k+1))×tan(3π/(2k+1))×.tan(k/(2k+1))=根号(2k+1)k为正整数)当k=1时,即tan(π/3)=根号3当k=2时,即tan(π/5)×tan(2π/5)=根号5. (4k^2+7k)+(-k^2+3k-1) 3×k×k-2k-1=-1.k等于 请问1^k+2^k+3^k+.+n^k=? 2/3π+2kπ与5/3π+2kπ的交集 函数y=根号下(2cosx+1)的定义域是( )A.[2kπ-π/3,2kπ+π/3] B.[2kπ-π/6,2kπ+π/6] C.[2kπ+π/3,2kπ+2π/3] D.[2kπ-2π/3,2kπ+2π/3]