Question

f(x)=√3(sin^2x-cos^2x)-2sinxcosx,设x∈[-π/3,π/3],求f(x)的单调递增区间。

答案是[π/12，π/3]求具体过程。谢谢~

Answer 1

f(x)=-√3cos2x-sin2x
=-2(sin2x*1/2+cos2x*√3/2)
=-2(sin2xcosπ/3+cos2xsinπ/3)
=-2sin(2x+π/3)
f(x)递增则sin(2x+π/3)递减
所以2kπ+π/2<=2x+π/3<=2kπ+3π/2
kπ+π/12<=x<=kπ+7π/12
x∈[-π/3,π/3]
显然k=-1和k=1时交集都是空集
k=0时，π/12<=x<=7π/12且x∈[-π/3,π/3]
所以是[π/12，π/3]

Answer 2

f(x)=√3(sin²x-cos²x)-2sinxcosx
=-√3cos(2x)-sin(2x)
=-2[(sin(2x)(1/2)+cos(2x)(√3/2)]
=-2sin(2x+π/3)
x∈[-π/3,π/3]，则-π/3≤2x+π/3≤π
当π/2≤2x+π/3≤π时，即π/12≤x≤π/3时，2sin(2x+π/3)单调递减，f(x)单调递增
递增区间为[π/12，π/3]。