$$e^{\sin x } - e^{- \sin x} - 4 = 0$$

Substitute $e^{\sin x} = y$: 
$$y - \frac{1}{y} - 4 = 0 \implies y^2 - 4y - 1 = 0$$

Solve for $y$: 
$$y = 2 \pm \sqrt{5}$$

$e^{\sin x}$ can't be negative:
$$\therefore y = 2 + \sqrt{5} \implies e^{\sin x} = 2 + \sqrt{5}$$
Differentiating both sides with respect to $x$:
$$\frac{d e^{\sin x}}{d x} = \frac{d (2 + \sqrt{5})}{d x} \implies \frac{d e^{\sin x}}{d \sin x} \times \frac{d \sin x}{d x} = 0 \implies e^{\sin x} \times \cos x = 0$$
If $e^{\sin x} = 0$ then $x$ has no solution. However, when $\cos x = 0$ then $x = 90^{\circ}$

The actual answer of the question states that there is no real solution of the equation. Then why am I getting the solution as $90^{\circ}$ too?