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)
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∵ 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
======以下答案可供参考======
供参考答案1:
函数y=log1/2[cos(x/3+π/4)]的单调递增区间就是cos(x/3+π/4）的减区间2kπ≤x/3+π/4≤(2k+1)π
且cos(x/3+π/4）大于0
2kπ＜x/3+π/4＜2kπ+π/2
6kπ-3/4π ＜ x＜6kπ+3/4π
供参考答案2:
因为1/2要求y=log1/2[cos(x/3+π/4)]的单调递增区间
所以就是求cos(x/3+π/4)的减区间
2kπ-π0
6kπ-3π/4