求函数y=log1/2[cos(x/3+π/4)]的单调递增区间.【1/2为底数】

∵ 1 ≤ (x/3+π/4)] ≤ 1
∴ 零和负数无对数
∴ 0＜ (x/3+π/4)] ≤ 1
0＜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/4≤ x ≤ 6kπ+9π/4 (k属于z)时,y= log（/2) h(x单调增