When does cosine multiplied by hyperbolic cosine equals one?

Question

What are the possible ways to solve $\cos(\alpha)\cdot\cosh(\alpha)=1$? I'm trying to get to an equation to find the roots of $\alpha$ here.

Any guidance on how to get there is much appreciated.

Answer 1

There are infinitely many (real) solutions. To get an idea of their nature render the equation as $\cosh(\alpha)=\sec(\alpha)$ and graph both functions. The secant function curves upwards more sharply than the hyperbolic cosine so that $\alpha=0$ is the only solution with absolute value less than ${3\pi}/2$. But for larger absolute values of $\alpha$ other solutions come in pairs satisfying $(2n-1/2)\pi < \alpha < (2n+1/2)\pi$ for each nonzero integer $n$. These nonzero solutions have no exact form in terms of elemmentary functions but are fairly easy to pinpoint numerically.

Answer 2

Probably worth mentioning that existence of the infinitely many real solutions is just application of the Intermediate Value Theorem with careful bookkeeping. However, other than $0,$ there is no reason to expect a closed form for any of the other solutions. Numerical approximations, yes.

There are infinitely many solutions. The easy one is $0.$ At any odd multiple of $\pi/2,$ cosine is $0.$

Let's begin with $ 3\pi/2 \approx 4.7123.$ Then $\cosh \frac{3 \pi}{2} \approx 55.66338,$ reciprocal 0.017965. Evidently the nearby root is $4.730040745,$ just a little bigger than $3\pi/2.$

Next $5 \pi/2 \approx 7.85398.$ The nearby root is $7.853204624,$ just a little smaller than $5\pi/2.$

Next $7 \pi/2 \approx 10.99557429.$ The nearby root is $10.99560784,$ just a little bigger than $7\pi/2.$

I need to do more work to figure why there is no solution to $$\cos x \cosh x - 1 = 0$$ near $x = \pi/2.$ Actually, Oscar has pointed out thet the single solution at $0$ should be thought of as a double root. Good thinking. Meanwhile, $$ \cos x \; \; \cosh x \; -1 \leq \; \frac{- x^4}{6} + \frac{ x^8}{2520} \; \; \; \mbox{for} \; \; \; |x| \leq 2.$$

I told it to divide the $y$ values by $50$ so it is possible to see more of the graph including the roots I mentioned up to 10.9956

Answer 3

As said in answers and comments, beside the trivial $x=0$, the roots of the equation $$\cos(x)\cosh(x)=1$$ are close to $\frac{(2k+1)\pi}2$ and, if $x$ is a root, $-x$ is also a root. So, let us just focus on the positive roots.

Since the first root is already large, we can make the approximation $\cosh(x)\approx \frac{e^x}2$. Moreover, around $x=\frac{(2k+1)\pi}2$, by Taylor $$\cos(x)\sim (-1)^{k+1}\left(x-\frac{(2k+1)\pi}2\right)$$ So, for the $k^{\text{th}}$ root, the equation can be approximated as $$(-1)^{k+1}\left(x_k-\frac{(2k+1)\pi}2\right)\frac{e^{x_k}}2=1$$ the solution of which being given in terms of Lambert function $$x_k=\frac{(2k+1)\pi}2+W\left( (-1)^{k+1}2 \exp\left(- \frac{(2k+1)\pi}2\right) \right)\tag1$$ For the first roots, this gives $$\left( \begin{array}{cc} k & x_k \\ 1 & 4.7300412 \\ 2 & 7.8532046 \\ 3 & 10.995608 \\ 4 & 14.137165 \\ 5 & 17.278760 \\ 6 & 20.420352 \\ 7 & 23.561945 \\ 8 & 26.703538 \\ 9 & 29.845130 \\ 10 & 32.986723 \end{array} \right)$$ which, for $k>1$, are exact for more than eight significant figures.

In fact for large $k$, the solution given by $(1)$ can be again approximated as $$x_k=\frac{(2k+1)\pi}2+ (-1)^{k+1}2 \exp\left(- \frac{(2k+1)\pi}2\right) \tag2$$