求函数y=log1/2[cos(x/3+π/4)]的单调递增区间.【1/2为底数】
问题描述：
求函数y=log1/2[cos(x/3+π/4)]的单调递增区间.【1/2为底数】
[6kπ-(3π)/4，6kπ+(3π)/4)(k属于z)
答
∵ 1 ≤ (x/3+π/4)] ≤ 1∴ 零和负数无对数∴ 0＜ (x/3+π/4)] ≤ 10＜1/2＜1∴y= log（/2) h(x)单调减即当cos(x/3+π/4)]在定义域内单调减时,y= log（/2) h(x）单调增即当2kπ ≤ x/3+π/4 ≤ 2kπ+π,即6kπ- 3π...