Question

求y=sin(-2χ+π/3)的单调区间及周期

Answer 1

解答：
y=sin（-2x+π/3 ）=-sin（2x-π/3 ）
(1)T=2π/2=π
(2)
增区间 ：即求y=sin(2x-π/3) 的减区间
2kπ+π/2≤2x-π/3≤2kπ+3π/2
2kπ+5π/6≤2x≤2kπ+11π/6
kπ+5π/12≤x≤kπ+11π/12
所以 函数y=sin（-2x+π/3 ）的单调递增区间是
【kπ+5π/12，kπ+11π/12】,k∈Z

减区间 ：即求y=sin(2x-π/3) 的增区间
2kπ-π/2≤2x-π/3≤2kπ+π/2
2kπ-π/6≤2x≤2kπ+5π/6
kπ-π/12≤x≤kπ+5π/12
所以 函数y=sin（-2x+π/3 ）的单调递减区间是
【kπ-π/12，kπ+5π/12】,k∈Z

Answer 2

2kπ-π/2<=-2x+π/3<=2kπ+π/2时单调递增
2kπ+π/2<=-2x+π/3<=2kπ+3π/2单调递减
T=π

Answer 3

T=2∏/|-2|=∏
单调增区间-π/2+2kπ≦-2x+π/3≦π/2+2kπ（k∈Z）
单调减区间π/2+2kπ≦-2x+π/3≦3π/2+2kπ（k∈Z)
自己解吧·

Answer 4

T=2π/w所以周期为π单调区间为（π/6+πk，7π/6+πk)