方程2sin(x+π/3)+2a-1=0在[0,π]上有两个不等的实根
sin(x+π/3)=1/2-a
因为x+π/3∈[π/3,4π/3],sin(x+π/3)∈[-√3/2,1]
所以当sin(x+π/3)∈[√3/2,1)时,x有两个不相等的实根
所以a∈(-1/2,(1-√3)/2]
方程2sin(x+π/3)+2a-1=0在[0,π]上有两个不等的实根
sin(x+π/3)=1/2-a
因为x+π/3∈[π/3,4π/3],sin(x+π/3)∈[-√3/2,1]
所以当sin(x+π/3)∈[√3/2,1)时,x有两个不相等的实根
所以a∈(-1/2,(1-√3)/2]