If $$0 \leqslant x \leqslant 2\pi ,$$   then number of roots of equation $${e^{\sin x}} - {e^{ - \sin x}} = 4{\text{ is}}$$

Solution :
The given equation can be written as
$$\eqalign{ & {e^{\sin x}} = 4 + \frac{1}{{{e^{\sin x}}}}\,\,\,.....\left( 1 \right) \cr & {\text{Now, }} - 1 \leqslant \sin x \leqslant 1\,{\text{and }}e < 3 \cr & \Rightarrow {e^{\sin x}} < 3 \cr & \Rightarrow {\text{Again as we always have }}\frac{1}{{{e^{\sin x}}}} > 0 \cr & \therefore 4 + \frac{1}{{{e^{\sin x}}}} > 4 \cr} $$
Thus the L.H.S. of $$\left( 1 \right) < 3$$  and R.H.S. of $$\left( 1 \right) > 4.$$
Hence there is no real values of $$x$$ which satisfy $$\left( 1 \right).$$
It follows that the given equation has no real solution.