已知实数a，b a²+ab+b²=1 t=ab–a²–b² 求t的范围 求大神帮忙

已知实数a，b a²+ab+b²=1 t=ab–a²–b² 求t的范围 求大神帮忙

解：
a²+ab+b²=1
a²+ab+¼b²+¾b²=1
(a+½b)²+[(√3/2)b]²=1
令a+½b=cosθ，(√3/2)b=sinθ
则a=cosθ-(1/√3)sinθ，b=(2/√3)sinθ
t=ab-a²-b²
=2ab-(a²+ab+b²)
=2ab-1
=2[cosθ-(1/√3)sinθ][(2/√3)sinθ]-1
=(2/√3)(2sinθcosθ)- (2/3)(2sin²θ)-1
=(2/√3)sin2θ - (2/3)(1-cos2θ)-1
=(4/√3)[½sin2θ +(√3/2)cos2θ] - 5/3
=(4√3/3)sin(2θ+ π/3) -5/3
-1≤sin(2θ+ π/3)≤1
-(4√3+5)/3≤(4√3/3)sin(2θ+ π/3) -5/3≤(4√3-5)/3
-(4√3+5)/3≤t≤(4√3-5)/3
t的取值范围为[-(4√3+5)/3，(4√3-5)/3]