题目内容
在[-π,π]内,函数
【答案】分析:由不等式2kπ-
≤x-
≤2kπ+
,求出x的取值范围,即得到函数
为增函数的区间.
解答:解:由函数
的解析式得,增区间满足 2kπ-
≤x-
≤2kπ+
,
∴2kπ-
≤x≤2kπ+
,k∈z,又x∈[-π,π],
∴-
≤x≤
,故增区间为[-
,
],
故答案为[-
,
].
点评:本题考查正弦函数的增区间的求法,列出不等式 2kπ-
≤x-
≤2kπ+
,是解题的关键.
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解答:解:由函数
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
∴2kπ-

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∴-
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故答案为[-
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点评:本题考查正弦函数的增区间的求法,列出不等式 2kπ-
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