帮做2道高1数学题

（1）用最大值区间做 令sin(2x+π/4)=1和sin(2x+π/4)=-1
对应的有2x+π/4=π/2和2x+π/4=-π/2 此区间为递增
解出 -3π/8＜x＜π/8 ＝＞ 2kπ-3π/8＜x＜2kπ+π/8 递增
同理可得到递减区间π/8＜x＜5π/8 ＝＞ 2kπ+π/8＜x＜2kπ+5π/8递减

（2)和第一题时一样的只不过是区间变了
cos(x+π/3)=1和cos(x+π/3)=-1
递减区间为 2kπ-π/3＜x＜2kπ+2π/3
递增区间为 2kπ-4π/3＜x＜2kπ-π/3

2kpai+pai/2<2x+pai/4<2kpai+3/2pai 递减