If the inequality sin2⁡x+acos⁡x+a2>1+cos⁡x holds for any x∈R then the largest negative integral value of a is

sin2⁡x+acos⁡x+a2>1+cos⁡xPutting X = 0, we have     a+a2>2 or     a2+a−2>0 or     (a+2)(a−1)>0 or     a<−2 or a>1Therefore, the largest negative integral value of a is -3.