If the equation cot4x−2cosec2x+a2=0  has at least one solution then, sum of all possible integral values of ‘a’ is equal to

We have  cot4x−2cosec2x+a2=0⇒cot4x−21+cot2x+a2=0⇒cot4x−2cot2x+a2−2=0⇒cot2x−12=3−a2To have at least one solution 3−a2≥0⇒a2−3≤0a∈−3,3⇒a=−1,0,1∴sum of all values of ‘a’ = 0