6.若tan(2x"-" π/6)≤1,则x的取值范围是　.

解析令z=2x-π/6,满足tan z≤1的z值是-π/2+kπ

　　即-π/2+kπ<2x-π/6≤π/4+kπ,k∈Z.

　　解得-π/6+1/2kπ

答案("-" π/6+1/2 kπ"," 5π/24+1/2 kπ],k∈Z

7.直线y=a与y=tan x的图像的相邻两个交点的距离是　　　　　.

解析由题意知,相邻两个交点间的距离即为一个周期的长度,故为π.

答案π

8.给出下列四个结论:

①sin-π/18>sin-π/10;②cos-25π/4>cos-17π/4;

③tan 5π/9>tan 17π/18;④tan π/5>sin π/5.

其中正确结论的序号是　　　　.

解析函数y=sin x是-π/2,0上的增函数,0>-π/18>-π/10>-π/2,所以sin-π/18>sin-π/10,①正确;cos-25π/4=cos-6π-π/4=cos π/4,cos-17π/4=cos-4π-π/4=cos π/4,所以cos-25π/4=cos-17π/4,②不正确;函数y=tan x是π/2,π上的增函数,π/2<5π/9<17π/18<π,所以tan 5π/9x>sin x,所以tan π/5>sin π/5,④正确.

答案①④

9.已知角α的终边上一点P的坐标为(-√3,y)(y≠0),且sin α=√2/4y.求tan α.

解由题意r2=x2+y2=3+y2,

　　由三角函数定义sin α=y/r=y/√(3+y^2 )=√2/4y,

　　∴y=±√5,∴tan α=y/("-" √3),即tan α=±√15/3.

10.利用函数图像解不等式-1≤tan x≤√3/3.

解作出函数y=tan x,x∈("-" π/2 "," π/2)的图像,如图所示.

　　观察图像可得:在区间("-" π/2 "," π/2)上,自变量x应满足-π/4≤x≤π/6.

　　由正切函数的周期性可知,不等式的解集为

　　{x├|"-" π/4+kπ≤x≤π/6+kπ"," k"∈" Z┤}.

11.求函数y=tan 2x的定义域、值域、单调区间,并作出它在区间[-π,π]内的图像.

解(1)要使函数y=tan 2x有意义,

　　只需2x≠π/2+kπ(k∈Z),即x≠π/4+kπ/2(k∈Z),

　　∴函数y=tan 2x的定义域为

　　{x├|x≠π/4+kπ/2 "," k"∈" Z}┤.

　　(2)设t=2x,由x≠π/4+kπ/2(k∈Z),知t≠π/2+kπ(k∈Z).∴y=tan t的值域为(-∞,+∞),

即y=tan 2x的值域为(-∞,+∞).