The correct option is D a∈[0,18)∪(2,∞)
22π/sin−1x−2(a+2)2π/sin−1x+8a<0
(2π/sin−1x−4)(2π/sin−1x−2a)<0
Now, 2π/sin−1x∈(0,14]∪[4,∞)
for 2π/sin−1x∈(0,14].We have
(2π/sin−1x−4)<0
∴ 22π/sin−1x−2a>0
or 0≤a<18
Similarly, for 2π/sin−1x∈[4,∞),a>2
So a∈[0,18)∪(2,∞)