A peculiar form

$\begin{aligned} \cosh 2 + i\sin \sqrt x & = \frac 1{e^2} \\ \frac {e^2+e^{-2}}2 + i \cdot \frac {i(e^{-\sqrt xi}-e^{\sqrt xi})}2 & = e^{-2} & \small \color{#3D99F6} \text{Multiplying both sides by }2 \\ e^2+e^{-2} - e^{-\sqrt xi}+e^{\sqrt xi} & = 2e^{-2} \\ e^{\sqrt xi} - e^{-\sqrt xi} & = e^{-2} - e^2 \\ \implies \sqrt xi & = - 2 & \small \color{#3D99F6} \text{Squaring both sides} \\ -x & = 4 & \small \color{#3D99F6} \text{Note that }i^2 = -1 \\ \implies x & = \boxed{-4} \end{aligned}$

We have that $\displaystyle\cosh{(2)}+i\sin{\left(\sqrt{x}\right)}=\dfrac{1}{e^2}$ . If we rewrite $\dfrac{1}{e^2} = e^{-2}$ as $\cos{\theta}+i\sin{\theta}$ , we see that $e^{-2} = e^{i(2i)}$ . From this we have $\theta = 2i$

$\cos{(2i)} = \dfrac{e^{i(2i)}+e^{-i(2i)}}{2} = \dfrac{e^{-2}+e^{2}}{2} =\ \cosh{(2)} \\ \therefore \cosh{(2)}+i\sin{(2i)} = \cos{(2i)}+i\sin{(2i)}$

Hence, we see that $\sqrt{x} = 2i \implies x = \boxed{-4}$