Find the exact value of sinh(ln 2) and solve cosh2(x) = 1 for x.

Using the property $e^{\ln a} = a$:

$$e^{\ln 2} = 2$$

Using the property $e^{-a} = \frac{1}{e^a}$:

$$e^{-\ln 2} = e^{\ln(2^{-1})} = 2^{-1} = \frac{1}{2}$$

First, simplify the numerator: $2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$.
Then, divide by 2:

$$\text{sinh}(\ln 2) = \frac{\frac{3}{2}}{2} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$$

The final answer is $\boxed{\frac{3}{4}}$.

:
We need to find $x$ given the equation $\cosh^2(x) = 1$.

Given $\cosh^2(x) = 1$, taking the square root of both sides gives:

$$\cosh(x) = \pm 1$$

The range of the hyperbolic cosine function, $\cosh(x)$, is $[1, \infty)$. This means $\cosh(x)$ can never be negative.
Therefore, we must have:

$$\cosh(x) = 1$$

$$\frac{e^x + e^{-x}}{2} = 1$$ $$e^x + e^{-x} = 2$$

$$e^x(e^x) + e^x(e^{-x}) = 2e^x$$ $$e^{2x} + e^0 = 2e^x$$ $$e^{2x} + 1 = 2e^x$$

Substitute $y$ into the equation.

$$y^2 - 2y + 1 = 0$$

This is a perfect square trinomial, which can be factored as $(y-1)^2 = 0$.

$$(y-1)^2 = 0$$ $$y-1 = 0$$ $$y = 1$$

$$e^x = 1$$

To solve for $x$, take the natural logarithm of both sides:

$$\ln(e^x) = \ln(1)$$ $$x = 0$$

The final answer is $\boxed{x=0}$.