题目内容
函数y=-3sin(2x-
)的单调递减区间为______.
| π |
| 6 |
令u=2x-
则函数y=3sinu的单调增区间为 [-
+2kπ,
+2kπ]k∈Z
由 -
+2kπ≤2x-
≤
+2kπ,得:
-
+kπ≤x≤
+kπk∈Z
∴函数y=3sin(2x-
)的单调增区间为:[-
+kπ,
+kπ]k∈Z
∴函数y=-3sin(2x-
)的单调递减区间为:[-
+kπ,
+kπ]k∈Z
故答案为:[-
+kπ,
+kπ]k∈Z
| π |
| 6 |
| π |
| 2 |
| π |
| 2 |
由 -
| π |
| 2 |
| π |
| 6 |
| π |
| 2 |
-
| π |
| 6 |
| π |
| 3 |
∴函数y=3sin(2x-
| π |
| 6 |
| π |
| 6 |
| π |
| 3 |
∴函数y=-3sin(2x-
| π |
| 6 |
| π |
| 6 |
| π |
| 3 |
故答案为:[-
| π |
| 6 |
| π |
| 3 |
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