Question

方程2sin(x+π/3)+2a-1=0在[0,π]上有两个不相等的实数根，则实数a的取值范围

Answer 1

2sin(x+π/3)+2a-1=0,在[0,π]上有两个不相等的实数根
就是1-2a=2sin(x+π/3)在[0,π]上左右函数图象有两个交点
x∈[0,π],(x+π/3)∈[π/3,4π/3],2sin(x+π/3)∈[-√3,2],
两图像有两个交点
于是1-2a∈(√3,2),
即a的取值范围是 -1/2<a<(1-√3)/2

Answer 2

2sin(x+π/3)+2a-1=0,就是2a=1-2sin(x+π/3)在[0,π]上的取值范围
x∈[0,π],(x+π/3)∈[π/3,4π/3],2sin(x+π/3)∈[－√3,2],于是1-2sin(x+π/3)∈[1,1+√3]
即a的取值范围是1/2≤a≤(1+√3)/2