y=lnx是增函数,根据复合函数规则(增增得增,增减得减)
要求y=lnsin(-2x+π/3)的单调递减区间
只要求y=sin(-2x+π/3)的单调递减区间
又sin[(π/3)-2x]=-sin(2x-π/3)
求函数f(x)的单调递减区间即求函数sin(2x-π/3)的单调递增区间,
2kπ-π/2≤2x-π/3≤2kπ+π/2,k∈Z.
kπ-π/12≤x≤kπ+5π/12,k∈Z.
∴所求的单调递减区间是[kπ-π/12,kπ+5π/12],k∈Z.
A.(kπ+5π/12,kπ+2π/3】B(kπ+π/6,kπ+5π/12】C(kπ+π/12,kπ+5π/12】D【kπ-π/12,kπ+π/6)(k∈Z)没有符合答案的选项················
完蛋,定义域没考虑,不好意思啊lnsin(-2x+π/3)还要考虑sin(-2x+π/3)>0即-sin(2x-π/3)>0即sin(2x-π/3)<0即-π+2kπ≤2x-π/3≤kπ得x∈[-2π/3+kπ,π/6+kπ](k∈z)综上选D