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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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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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)
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)

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)
∵ 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

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

因为1/2<1所以y=log1/2x为减函数
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