Question

求函数y=根号3sinx+cosx的周期，单调递区间，最值

Answer 1

y=2sin(x+π/6)
周期是2π，最大值是2，最小值是-2
单调递增区间（2nπ-π/3,2nπ+2π/3)
递减区间是（2nπ+2π/3，2nπ+5π/3)

Answer 2

y=根号3sinx
cosx
=2(sinx*cosπ/6
sinπ/6cosx)
=2sin(x＋π/6）
由y=sinx的单调性知，已知函数在：-π/2
2kπ<=x＋π/6<=3π/2
2kπ即：-2π/3
2kπ<=x<=4π/3
2kπ上增
在4π/3
2kπ<=x<=7π/3
2kπ上减，k为整数。