求函数y=log1/2sin(π/3-2x)的单调递增区间

题目

求函数y=log1/2sin(π/3-2x)的单调递增区间

答案

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...