使函数y=3sin(-2x-π6)为增函数的区间为(  ) A.[0,5π12] B.[2π3,11π12] C.[π6,11π12] D.[π6,2π3]

问题描述:

使函数y=3sin(-2x-

π
6
)为增函数的区间为(  )
A. [0,
12
]
B. [
3
11π
12
]
C. [
π
6
11π
12
]
D. [
π
6
3
]

∵y=3sin(-2x-π6)=-3sin(2x+π6),由复合函数的性质得,函数y=3sin(2x+π6)的单调递减区间就是y=3sin(-2x-π6)的单调递增区间,∴由2kπ+π2≤2x+π6≤2kπ+3π2(k∈Z)得:kπ+π6≤x≤kπ+2π3(k∈Z)...