解三角不等式,复合函数单调区间练习
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解三角不等式
(1)cos x ≥
[2,2],66k k k Z ππππ-+∈
(2)cos x ≤
11[2,2],66k k k Z ππππ++
∈
(3)sin x ≥
2[2,2],33k k k Z ππππ++
∈
(4)sin 2
x ≤
4[2,2],33k k k Z ππππ-+∈
(5)tan x ≥ [,),32k k k Z π
π
ππ++∈
(6)tan x ≤ (,],23
k k k Z ππ
ππ-+∈ 求定义域
(1)()f x = {|222,}2
x k x k x k k Z π
ππππ≤<+=+∈或
(2)()f x = {|22}2
x k x k k Z π
ππ≤<+
∈
(3)()lg(cos )f x x = {|22,}26x k x k k Z ππππ-<≤+∈
(4) ()lg(sin )f x x = {|22,}3x k x k k Z ππππ+≤<+∈
(5) ()f x =
53{|22,}42x k x k k Z ππππ+<<+∈
(6)()f x =
43(2,2)(2,2),3232k k k k k Z ππππππππ++⋃+
+∈
(7) lg(tan ()x f x =
73(2,2)(2,2),3262k k k k k Z ππππππππ++⋃++∈
(8) ()f x =
(,),42
k k k Z π
π
ππ+
+∈
(9) ()f x =
33(2,2)(2,2),3242k k k k k Z ππππππππ++⋃+
+∈
(10) lg(tan ()x f x = 73(2,2)(2,2),3262k k k k k Z ππππππππ++⋃++∈
求单调区间
(1)()lg(sin )f x x =
增：(2,2],2
k k k Z π
ππ+∈ 减：[2,2),2
k k k Z π
πππ+
+∈
(2)12
()log (sin )f x x =
增：[2,2),2k k k Z π
πππ+
+∈ 减：(2,2],2
k k k Z π
ππ+∈
(3)()lg(sin 2)f x x =
增：(,],4
k k k Z π
ππ+
∈ 减：[,),42
k k k Z π
π
ππ+
+∈ (4)12
()log (sin 2)f x x =
增：[,),42k k k Z π
πππ+
+∈ 减：(,],4
k k k Z π
ππ+∈
(5)()lg(cos )f x x =
增：(2,2],2k k k Z πππ-
∈ 减：[2,2),2
k k k Z π
ππ+∈ (6) 12
()log (cos )f x x =
增：[2,2),2
k k k Z π
ππ+∈ 减：(2,2],2
k k k Z π
ππ-
∈
(7) ()lg(tan )f x x =
增：(,),2
k k k Z π
ππ+
∈ 无减区间