(1)单调增区间(2kπ-π/2,2kπ+3π/2)
(2)单调增区间【kπ+3π/8,kπ+7π/8】
y=tanx的单调增区间是(kπ-π/2,kπ+π/2),那么求这个函数的递增区间就有:
x/2-π/4在范围(kπ-π/2,kπ+π/2)内,解不等式得到(2kπ-π/2,2kπ+3π/2).
第二题先把负号提出来成为y=-sin(2x-π/4)然后找sin(2x-π/4)的递减区间即可,那么y=sinx的递减区间是【2kπ+π/2,2kπ+3π/2】,那么2x-π/4的取值范围是【2kπ+π/2,2kπ+3π/2】解得x的范围【kπ+3π/8,kπ+7π/8】.