If esinx−e−sinx=a has alteast one real solution, then
esinx−e−sinx=a..............(1)
Let esinx=t⇒t∈[1e,e]
∴(1)⇒t−1t−a=t2−at−1=0
Discriminant =a2+4>0∀ a ∈ R.
Also product of roots =-1
i.e. both the roots can't belong to [1e,e]
Let it have only one root in [1e,e], then
(e2−ae−1)(1e2−ae−1)≤0=[(e2−1)−ae][(e2−1)+ae]≥0⇒|a|≤e2−1e