Question

设函数f（x）=sinx-根号3cosx(x属于[-π，0]的单调递增区间是

Answer 1

f(x)=sinx-√3×cosx=2(1/2×sinx-√3/2×cosx)=2(sinx × cos(π/3)-cosx × sin(π/3))
=2sin(x-π/3),x∈[-π,0]
令y=x-π/3，则
f(y+π/3)=2siny,y∈[-4π/3,-π/3],
因为正弦函数的单调递增区间在1和4象限，而y区间在1和4象限的区域为[-π/2,-π/3],
将x=y+π/3代入，则
f(x)的单调递增区间为[-π/2+π/3,-π/3+π/3]，即[-π/6,0]。

Answer 2

F(X)=sinx-根号3cosx
=2sin(x-pai\3)
x属于(-π\2+2kπ,2kπ)
所以x-π\3π(-5π\6+2kπ,-π+2kπ)

Answer 3

[-（2/3）π，0]

Answer 4

[-π/6 ,0]

Answer 5

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