Question

一道三角函数问题，详见图片

Answer 1

f(x)=2[1/2 sinx-根号3/2 cosx]=2sin(x-π/3),    

-π<=x<=0,    

2kπ-π/2<=x-π/3<=2kπ+π/2   时为递增 

2kπ-π/6<=x<=2kπ+5π/6  (k为整数)

因为已知-π<=x<=0,  

所以当k=0时,   -π/6<=x<=5π/6 与上式取交集

得-π/6<=x<=0 

Answer 2

主要是恒等变换，提出一个2来，将后面的式子变形运用三角公式。

注意前面的负号

Answer 3

f(x)=2(sinxcosπ/3-cosxsinπ/3)=2sin(x-π/3)
递增区间为2kπ-π/2≤x-π/3≤2kπ+π/2
结合-π≤x≤0得递增区间-π/6≤x≤0