If we have the determinant of matrix $\sinh x \cosh x=0$
Then $\sinh(x)=0$ or $\cosh(x)=0$
If $\sinh(x)=0$, then $x=0, \pi, 2\pi, 3\pi$
And $\cosh(x)=0$ then $x=\pi/2, 3\pi/2, 5\pi/2$
Is that correct or not? How can we find the value of $x$ ?
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$\begingroup$The hyperbolic functions are quite different from the circular ones. For one thing, they are not periodic.
For your equation, the double-"angle" formula can be used:
$\sinh x \cosh x = 0$
$\frac 12 \sinh 2x = 0$
$\sinh 2x = 0$
The only solution to that is $2x = 0 \implies x = 0$.
Alternatively, you can simply observe that $\cosh x$ is always non-zero, and the only solution comes from $\sinh x = 0$.
Updated: in the complex numbers, $2x = k\pi i \implies x = \frac 12 k\pi i, k \in \mathbb{Z}$. See my comment below.
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