If the equation cot4x−2cosec2x+a2=0 has at least one solution, then the sum of all possible integral values of a is equal to
4
3
2
0
cot4x−21+cot2x+a2=0 or cot4x−2cot2x+a2−2=0 or cot2x−12=3−a2
To have at least one solution 3−a2≥0
a2−3≤0a∈[−3,3]
Integral values are -1, 0, 1; therefore, the sum is 0.