求出函数f(x)=sin(x+π/3)+√3sin(x-π/6)的单调增区间.
问题描述:
求出函数f(x)=sin(x+π/3)+√3sin(x-π/6)的单调增区间.
答
将后面的正弦化为余弦,然后在式子两边提出2整理成y=1/2(sin(x+2π/3))后易得.
求出函数f(x)=sin(x+π/3)+√3sin(x-π/6)的单调增区间.
将后面的正弦化为余弦,然后在式子两边提出2整理成y=1/2(sin(x+2π/3))后易得.