Answer
$(\ln(1+\sqrt{2}),\sqrt{2})$
Work Step by Step
The slope of the tangent line to the graph of the function $f(x)=\cosh(x)$ at the point $(x,f(x))$ is:
$$f'(x)=(\cosh(x))'=\sinh(x)=\frac{e^{x}-e^{-x}}{2}$$ $$1=\frac{e^{x}-e^{-x}}{2}$$ $$2=e^{x}-e^{-x}$$ $$2=e^{x}-\frac{1}{e^{x}}$$ $$2=\frac{e^{x} \cdot e^{x}-1}{e^{x}}$$ $$2=\frac{(e^{x})^{2}-1}{e^{x}}$$ $$2e^{x}=(e^{x})^{2}-1$$ $$0=(e^{x})^{2}-2e^{x}-1$$
Let $t=e^{x}$ so the equation becomes:
$$0=t^{2}-2t-1$$
The solution of the above quadratic equation is: $$t=\frac{2+2\sqrt{2}}{2},t=\frac{2-2\sqrt{2}}{2}$$ $$t=1+\sqrt{2},t=1-\sqrt{2}$$ $$e^{x}=1+\sqrt{2},e^{x}=1-\sqrt{2}$$
Since $e^{x} \gt 0$ for all $x$ in $\mathbb R$, the equation $e^{x}=1-\sqrt{2}$ should be discarded, so:
$$e^{x}=1+\sqrt{2}$$ $$x=\ln(1+\sqrt{2})$$
So the y-coordinate of $x=\ln(1+\sqrt{2})$ is: $$f(x)=\cosh(x)=\frac{e^{x}+e^{-x}}{2}$$ $$f(\ln(1+\sqrt{2}))=\cosh(\ln(1+\sqrt{2}))=\frac{e^{\ln(1+\sqrt{2})}+e^{-(\ln(1+\sqrt{2}))}}{2}=\frac{1+\sqrt{2}+\frac{1}{(1+\sqrt{2})}}{2}=\sqrt{2}$$ The point is: $$(\ln(1+\sqrt{2}),\sqrt{2})$$
