If the equation cot4x−2cosec2x+a2=0 has at least one solution then, sum of all possible integral values of ‘a’ is equal to
4
3
2
0
We have cot4x−2cosec2x+a2=0
⇒cot4x−21+cot2x+a2=0
⇒cot4x−2cot2x+a2−2=0
⇒cot2x−12=3−a2
To have at least one solution 3−a2≥0
⇒a2−3≤0
a∈−3,3
⇒a=−1,0,1
∴sum of all values of ‘a’ = 0