函数f(x)=sin2x·sin(π/6)-cos2x·cos(5π/6)在[-π/2,π/2]上的单调递增区间为?

函数f(x)=sin2x·sin(π/6)-cos2x·cos(5π/6)在[-π/2,π/2]上的单调递增区间为?
函数f(x)=sin2x·sin(π/6)-cos2x·cos(5π/6)在[-π/2,π/2]上的单调递增区间为?

函数f(x)=sin2x·sin(π/6)-cos2x·cos(5π/6)在[-π/2,π/2]上的单调递增区间为?
直接得到：F（x）=sin（2X+π/3）,由于是正弦函数,所以-π/2=