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求解log(1/2)cos(x+π/6)≤1/2,x的取值范围
人气:232 ℃ 时间:2020-04-19 14:45:21
解答
log(1/2)cos(x+π/6)≤1/2=log(1/2)[(1/2)^1/2]=log(1/2)(√2/2)
log(1/2)x递减
所以cos(x+π/6)≥√2/2=cos(2kπ±π/4)
所以2kπ-π/4≤x+π/6≤2kπ+π/4
2kπ-5π/12≤x≤2kπ+π/12
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