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求函数y=log1/2sin(π/3-2x)的单调递增区间
人气:275 ℃ 时间:2019-10-10 05:48:07
解答
y=log(1/2) sin(π/3-2x)∵0<1/2<1∴y的单调递增区间,即为sin(π/3-2x)单调递减区间令A=sin(π/3-2x)=-sin[2x-(π/3)]只要求得sin[2x-(π/3)]单调递增区间即可,得:kπ-(π/2)≤x-(π/3)≤kπ+(π/2)kπ-(π/6)≤x...
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