Question

求f(x)=2sin(2x+π/6)在区间[0，π]的单调递增区间，

Answer 1

f(x)=2sin(2x+π/6)
设t=2x+π/6，t∈[π/6，2π+π/6]
f(t)=2sint
f(t)的单调递增区间为：
t∈[π/6，π/2]或t∈[3π/2，2π+π/6]
此时：
x∈[0，π/6]或x∈[2π/3，π]。
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Answer 2

y=sinx,在[-π/2
+2kπ，π/2
+2kπ]递增
所以-π/2
+2kπ≤2x+π/6≤π/2
+2kπ，即-π/3+kπ≤x≤π/6
+kπ
当k=0时，0<x≤π/6;
当k=1时，2π/3≤x<π
所以，(0,π/6]u[2π/3,π)