f(x）=2sin(2x-pai/3)+1

题目

f(x）=2sin(2x-pai/3)+1
求f(x）>0的集合
f(x)在[0,pai]上的递增区间和递减区间

答案

sin(2x-π/3)>-1/2=sin(2kπ-π/6)=sin(2kπ+7π/6)
所以2kπ-π/6<2x-π/3<2kπ+7π/6
2kπ+π/6<2x<2kπ+3π/2
kπ+π/12x∈{x|kπ+π/120-π/3<2x-π/3<5π/3
sin增区间是(2kπ-π/2,2kπ+π/2)
所以-π/3<2x-π/3<π/2和3π/2<2x-π/3<5π/3是增区间
-π/3<2x-π/3<π/2
0<2x<5π/6
03π/2<2x-π/3<5π/3
11π/12所以增区间(0,5π/6)(11π/12,π)
减区间(5π/6,11π/12)