Question

高二数学文科，有图，求详细过程，因为基础差

Answer 1

解函数y=sin(2x+π/3)
知当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z，y是增函数
即当2kπ-5π/6≤2x≤2kπ+π/6,k属于Z，y是增函数
即当kπ-5π/12≤2x≤kπ+π/12,k属于Z，y是增函数
故函数和斗的增区间卖搜为[kπ-5π/12，kπ+π/12],k属中棚历于Z。

Answer 2

∵ y=sin(2x+π/3),
∴由2kπ−π/2≤段碰2x+π/3≤2kπ+π/2, k∈Z.
得kπ-5π/12≤x≤kπ+π/12, k∈Z．
∴当k=0时, 递增区间为[0, π/12],
当k=1时, 递增区间为[7π/12, π],
即洞燃答在[0，π]内的单调增区间是[0, π/12]和纳慧[7π/12, π]．

Answer 3

由2kπ-π/2≦2x+π/3≦2kπ+π/2
得2kπ-5π/6≦2x≦2kπ+π/6
得kπ-5π/12≦数孝x≦kπ+π/薯冲稿12;
即单增区间判唤为：[kπ-5π/12，kπ+π/12](k∈Z）

Answer 4

单调递含顷败增
2kπ -π/2 ≤2x+π/谈颤3≤乎闷 2kπ +π/2
2kπ -5π/6 ≤2x ≤ 2kπ +π/6
kπ -5π/12 ≤x ≤ kπ +π/12