Question

f(x)=根号3（sin²x-cos²x）-2sinxcosx+1 如何化简呢？

（1）求f(x)的最小正周期 (2）设x属于[-π/3,π/3],求f(x)的值域和单调递增区间。 本人第一次在这里提问题呢~~~~谢谢各位大虾的回答，，，求详细过程~~~xiexie^^

Answer 1

解：f(x)=√3(sin²x-cos²x)-2sinxcosx+1
=-√3(cos²x-sin²x)-sin2x+1
=-√3cos2x-sin2x+1
=-2(sinπ/3cos2x+cosπ/3sin2x)+1
=-2sin(2x+π/3)+1
(1)最小正周期T=2 π/2=π
(2)-π/3<=x<=π/3有 -π/3<=2x+π/3<=π
据函数图象有 当2x+π/3=-π/3时，f(x)取得最大值√3+1，当2x+π/3=π/2时，f(x)取得最小值-1
所以f(x)值域为【-1，√3+1】
由图象有-π/3<=2x+π/3<=π/2是单调减函数，计算有x的值为[-π/3,π/12]
由图象有π/2<=2x+π/3<=π是单调增函数，计算有x的值为[π/12，π/3]

Answer 2

题目不清楚，)=根号3（sin²x-cos²x）-2sinxcosx+1
最小正周期应该是π

Answer 3

f(x)=-根号3cos2x-sin2x+1=-sin(2x+π/3)+1

追问

看不懂~~~