f(x)= sin2x+2√3cos²x-√3-2a= sin2x+√3cos2x-2a=2sin(2x+π/3)-2a,
因为x∈(0,π/2),所以,(2x+π/3)∈(π/3,4π/3),
由f(x)在(0,π/2)内唯一的零点,由图知(2x+π/3)∈[2π/3,4π/3),或(2x+π/3)=π/2,
此时,a=sin(2x+π/3)∈(-√3/2,√3/2]或a=1.
即a的取值范围是{a| -√3/2
f(x)= sin2x+2√3cos²x-√3-2a= sin2x+√3cos2x-2a=2sin(2x+π/3)-2a,
因为x∈(0,π/2),所以,(2x+π/3)∈(π/3,4π/3),
由f(x)在(0,π/2)内唯一的零点,由图知(2x+π/3)∈[2π/3,4π/3),或(2x+π/3)=π/2,
此时,a=sin(2x+π/3)∈(-√3/2,√3/2]或a=1.
即a的取值范围是{a| -√3/2