急急急!!!求高人解答

2024-12-22 01:56:39
推荐回答(1个)
回答1:

函数f(x)=根号3(sinx的平方-cosx的平方)-2sinxcosx
=-根号3cos2x-sin2x
=-2 (根号3/2cos2x+1/2sin2x)
=-2(sinπ/3cos2x+cosπ/3sin2x)
=-2sin(2x+π/3)
最小正周期=2π/2=π
当2kπ+π/2≤2x+π/3≤2kπ+3π/2时递增
即 2kπ+π/6≤2x≤2kπ+7π/6
kπ+π/12≤x≤kπ+7π/12, k∈Z
即增区间为
[kπ+π/12,kπ+7π/12],k∈Z