求函数y=sin(2x+π/4)x属于[0,π]的单调递增区间

令该函数单调递增则其导数必为正数,可有
y' = (3sin(2x+π/4))' = 3cos(2x+π/4)(2x+π/4)' =6cos(2x+π/4) > 0
令 t = 2x+π/4 则 t ∈ (-π/2,π/2) 时 y' 恒大于0,于是有
π/2 < t = 2x+π/4 < -π/2 化简之有 -3π/8 < x < π/8--------①
根据题意 0 ≦ x ≦ π--------②,联立①②求交集可得到
0 < x < π/8 即为函数y=3sin(2x+π/4)x属于[0,π]的单调递增区间.