y=log底数是2 sin(-2x+三分之π)+1的单调增区间

函数y=log（2） x是单调增函数
所以
函数y=log（2）[ sin(-2x+π/3)+1]函数的单调性由sin(-2x+π/3)+1决定
函数y=sinx的单调区间是
[2Kπ-π/2,2Kπ+π/2]单调递增
[2Kπ+π/2,2Kπ+3π/2]单调递减
2Kπ-π/2≤-2x+π/3≤2Kπ+π/2
2Kπ-5π/6≤-2x≤2Kπ+π/6
-Kπ+5π/12≥x≥-Kπ-π/12
所以Kπ-π/12≤x≤Kπ+5π/12时,函数单调递增
同理可得Kπ+5π/12≤x≤Kπ+11π/12时,函数单调递减