求函数y=log1/2sin(π/3-2x)的单调递增区间
2个回答

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≤kπ+(5π/6)

且,x-(π/3)≠kπ,x≠kπ+(π/3)

单调递增:kπ-(π/6)≤x≤kπ+(5π/6) 且x≠kπ+(π/3) ,k∈整数