问题标题：
求.函数y=3sin(π/6-3x),x∈[-π/2,π/2]的单调递增区间
问题描述：
求.函数y=3sin(π/6-3x),x∈[-π/2,π/2]的单调递增区间
谈正回答：
因x∈[-π/2,π/2],故π/6-3x∈[-4π/3,5π/3]
由π/6-3x∈[-4π/3,-π/2],得x∈[2π/9,π/2]
由π/6-3x∈[π/2,3π/2],得x∈[-4π/9,-π/9]
故,单调递增区间为[2π/9,π/2]和[-4π/9,-π/9]
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