Why is my solution of $e^{\sin x } - e^{- \sin x} - 4 = 0$ wrong? $$e^{\sin x } - e^{- \sin x} - 4 = 0$$
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?

Let $e^{\sin x} = 2 + \sqrt{5}$. Then, $\sin{x}=\ln{(2+\sqrt{5})}>\ln{3}>1$, which isn't possible (since $\sin{x}\in[-1,1]$ for every $x$). Therefore, there is no solution.
