求函数y=sin(3x+π/6),x∈[0,π/2]的单调递减区间及函数y=cos(x+π/3)的在[0,2π]上的单调

①正弦函数的减区间为[2kπ+π/2,2kπ+3π/2]
所以2kπ+π/2≤3x+π/6≤2kπ+3π/2
2kπ+π/3≤3x≤2kπ+4π/3
∴2kπ/3+π/9≤x≤2kπ+4π/9,k∈Z
∵定义域是[0,π/2]
∴π/9≤x≤4π/9
所以单调减区间是[π/9,4π/9]
②余弦函数的增区间是[2kπ-π,2kπ]
∴2kπ-π≤x+π/3≤2kπ
∴2kπ-4π/3≤x≤2kπ-π/3
当k=1时,2π/3≤x≤5π/3
∴增区间是[2π/3,5π/3]