Question

1.求函数y=sin(2x-π/3)的对称中心

2.求函数y=sin(π/3-2x)的单调递增区间、对称轴

Answer 1

1、函数y=sin(2x-π/3)的对称中心
就是y=sin(2x-π/3)=0
的解
即：2x-π/3=kπ
所以
x=kπ/2+π/6
2、增区间满足
kπ-π/2<=π/3-2x<=kπ+π/2
即：X∈【kπ/2-
π/12
kπ/2+5π/12】
对称轴
就是y=sin(2x-π/3)=1或-1
的解
即：π/3-2x=kπ+π/2
x=kπ/2-π/12

Answer 2

1、函数y=sin(2x-π/3)的对称中心 就是y=sin(2x-π/3)=0 的解 即：2x-π/3=kπ 所以 x=kπ/2+π/6
2、增区间满足 kπ-π/2<=π/3-2x<=kπ+π/2 即：X∈【kπ/2- π/12 kπ/2+5π/12】
对称轴 就是y=sin(2x-π/3)=1或-1 的解 即：π/3-2x=kπ+π/2 x=kπ/2-π/12

Answer 3

y=sin(2x-π/3) = 2sin(x-π/6) x-π/6=Kπ+π/2 x=Kπ+2/3π 对称轴
y=sin(π/3-2x) = 2sin(π/6-x) π/6-x=Kπ+π/2 x=-Kπ-1/3π 对称轴
单调递增区间：2Kπ-π/2《π/3-2x《2Kπ+π/2 -2Kπ+π/2》2x-π/3》-2Kπ-π/2
2Kπ+5π/6》2x》-2Kπ-π/6
-Kπ-π/12《x《Kπ+5π/12

Answer 4

1、2x-π/3=kπ x=kπ/2+π/6
2、增区间 kπ-π/2<π/3-2x<kπ+π/2 kπ/2- π/12< x<kπ/2+5π/12
对称轴 π/3-2x=kπ+π/2 x=-kπ/2-π/12