函数y=sin(-x+π3)(x∈[0,2π]的单调减区间是______.-数学

题目简介

函数y=sin(-x+π3)(x∈[0,2π]的单调减区间是______.-数学

题目详情

函数y=sin(-x+
π
3
)(x∈[0,2π]
的单调减区间是______.
题型:填空题难度:中档来源:不详

答案

∵y=sin(-x+class="stub"π
3

=-sin(x-class="stub"π
3
),(x∈[0,2π])
∴函数y=sin(-x+class="stub"π
3
)(x∈[0,2π])的单调减区间为y=sin(x-class="stub"π
3
)的单调增区间.
∴由2kπ-class="stub"π
2
≤x-class="stub"π
3
≤2kπ+class="stub"π
2
(k∈Z)得:
2kπ-class="stub"π
6
≤x≤2kπ+class="stub"5π
6
,k∈Z,
又x∈[0,2π],
∴0≤x≤class="stub"5π
6
class="stub"11π
6
≤x≤2π.
故答案为:[0,class="stub"5π
6
],[class="stub"11π
6
,2π].

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