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

Question

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

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?

Answer 1

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.

Answer 2

Hint :

You have $$e^{sin(x)}=2+\sqrt{5}$$ implying $$sin(x)=1.443\cdots$$

by taking $\ln$ on both sides.

Take it from here.

Answer 3

$e^{\sin x} = 2 + \sqrt{5}$ has no solutions!

In fact it is equivalent to $\sin(x)=\ln(2 + \sqrt{5})>\ln(4)>1$ which is impossible because $\sin(x)\in[-1,1]$.

Answer 4

By the way, your equation is equivalent to $$\sinh(\sin(x))=2\iff \sin(x)=\text{arcsinh}(2)>1.$$

Answer 5

You say '$e^{sinx}$ cannot be negative' - WRONG! It certainly CAN be negative if $x$ is not real! Therefore your excluding the second value of $y$ is erroneous.