$$\cos(\sqrt{n^2-1}) = \frac{1}{n}$$
I was wondering if there existed integer solutions to to this equation apart from n=1. I've thought that there are probably no more solutions, because RHS is rational and I believe that LHS is not rational in general. I've found theorems about the output of the cosine when the argument is rational, but not for when it is irrational. (Niven's theorem).
How could this be solved?
What are the integer solutions to $\cos(\sqrt{n^2-1}) = \frac{1}{n}$
diophantine equationsirrational-numberstrigonometry
Related Solutions
First of all, your example equation doesn't correspond to your problem: your $b$ in $\sin(ax+b)=\sin(x)$ is not non-zero - differently from what you have prescribed by your constraints.
The conditions of your point 1 ("$x$ is a rational multiple of $\pi$") can only be met if $b=0$, but your constraints prohibit zero $b$.
Considering that $\sin$ and $\arcsin$ have the exceptional set $\{0\}$ (see below), the conditions of your point 2 can only be met if $b=0$, but your constraints prohibit zero $b$.
Your point 3 is just an assumption from you.
Clearly, your thoughts are far from complete.
$\ $
Let me present a more systematic way.
"Closed form" means expressions of allowed functions. If an equation is solvable in closed form depends therefore on the functions you allow.
What you are looking for are solutions that are closed-form numbers.
Clearly, answers for closed-form solutions can be given only for single given classes of functions. Clearly, integers, rational numbers and algebraic numbers are usually allowed as closed-form numbers.
Let's consider an equation $\text{trig}(ax+b)=P(x)$, with $a, b \in \mathbb{Q}$, $a\neq 0$, where $\text{trig}$ is one of the trigonometric functions and $P\in\mathbb{Q}(x)$ is not constant. The trigonometric functions are the functions $\sin$, $\cos$, $\tan$, $\cot$, $\sec$ and $\csc$.
We can treat this kind of equations in the same way for all trigonometric functions because the function term of a trigonometric function can be represented as a non-conctant rational expression of $e^{ix}$ over the algebraic numbers, e.g.: $\sin(x)=R(e^{ix})=-\frac{1}{2}i\left(e^{ix}-e^{-ix}\right)$, with non-constant $R\in Quot(\overline{\mathbb{Q}}(x))$. Let's call this kind of presentation of the trigonometric functions their exponential representation.
1.) Algebraic numbers
Let's allow the algebraic numbers as closed form.
Let $x$ be algebraic. $P(x)$ is algebraic then.
For the trigonometric functions, Lindemann-Weierstrass theorem helps to determine the exceptional sets. See my answer at On the behavior of transcendental functions. Because the only algebraic sets of trigonometric functions are $\{\}$ or $\{0\}$, $\text{trig}(ax+b)$ can for algebraic $x$ be algebraic only if $ax+b=0$ $\ \ $ ($x=-\frac{b}{a}$).
Let $P(x)$ a polynomial as defined above that has $-\frac{b}{a}$ as a solution. The trigonometric functions have the following exceptional points. $\sin,\tan\colon (0,0)$; $\cos,\sec\colon (0,1)$. $\cot$ and $\csc$ don't have an exceptional point. Of the equations of the kind prescribed in the question, exactly the following have an algebraic solution therefore.
$$\sin(ax+b)=P(x)$$
$$\tan(ax+b)=P(x)$$
$$\cos(ax+b)=1+P(x)$$
$$\sec(ax+b)=1+P(x)$$
2.) Elementary inverses, Elementary numbers
Let's now allow the elementary functions as closed form.
The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms.
a) Solvability of equations by closed-form expressions by rearranging the equation is related to the question of existence of inverses of the functions which are contained in the equation.
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in [Ritt 1925], that is also proved in [Risch 1979], answers which kinds of elementary functions can have an inverse which is an elementary function. We can extend this theorem from the inverses of the bijective elementary functions to the partial inverses of arbitrary elementary functions by treating restrictions of the non-bijective elementary functions by restricting their domains.
In some cases, you can also take the method of [Rosenlicht 1969].
As mentioned above, for each trigonometric function $\text{trig}$, there is an exponential presentation of the equation $\text{trig}(ax+b)=P(x)$: $\ R(e^{i(ax+b)})=P(x)$, with non-constant $R\in Quot(\overline{\mathbb{Q}}(x))$. This equation can be transformed to an equation $P_{1}(e^{i(ax+b)},x)=0$, where $P_{1}$ is a non-constant binary polynomial over the algebraic numbers because $\text{trdeg}_{\overline{\mathbb{Q}}}(e^{(ax+b)i},x)>0$. The left-hand side of this equation is an elementary function of $x$. Let's name it $f$. If the equation is solvable by applying only elementary functions to the equation that we can read from the equation, the inverse of $f$ has to be an elementary function. A conclusion of the theorem of Ritt in [Ritt 1925] is: If $f$ and its inverse are both elementary, a function term for $f$ in exp-ln representation must exist that doesn't contain an multiary algebraic function. But for each trigonometric function $\text{trig}$, $P_1$ is multiary. Therefore we cannot solve the equations by rearranging them by applying only elementary functions that we can read from the equation.
b) Solving an equation by closed-form expressions means to find solutions that are closed-form numbers. A closed-form number is a number that is generated by applying a finite number of closed-form functions to a rational number. [Chow 1999] asked this problem. One first and simple solution of the problem is Lin's theorem in [Lin 1983]:
"If Schanuel's conjecture is true and $f(X,Y)\in\overline{\mathbb{Q}}[X,y]$" is an irreducible polynomial involving both $X$ and $Y$ and $f(\alpha,exp(\alpha))=0$ for some non-zero $\alpha\in\mathbb{C}$, then $\alpha$ is not in EL."
$EL$ are the elementary numbers. [Chow 1999] treats the explicit elementary numbers.
For each trigonometric function $\text{trig}$, the equation $P_{1}(e^{(ax+b)i},x)=0$ (see section a) is irreducible and therefore of the kind of Lin's theorem. That means, the equations don't have a non-zero solution that is an elementary number.
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
3.) Lambert W and Elementary functions
Let's now allow the algebraic functions and Lambert W as closed form.
We use the exp-ln representation of the trigonometric functions (See e.g. the Wikipedia articles for the single functions or [Abramowitz/Stegun 1970].):
$$\sin(x)=-\frac{1}{2}ie^{ix}+\frac{1}{2}ie^{-ix},\ \cos(x)=\frac{1}{2}e^{ix}+\frac{1}{2}e^{-ix},\ \tan(x)=-i\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}},$$ $$\cot(x)=i\frac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}},\ \sec(x)=\frac{2}{e^{ix}+e^{-ix}},\ \csc(x)=\frac{2i}{e^{ix}-e^{-ix}}.$$
For each trigonometric function $\text{trig}$, the equation $\text{trig}(ax+b)=P(x)$ can be transformed to a quadratic polynomial equation of $e^{iax}$:
$$p_0(x)+p_1(x)e^{iax}+p_2(x)\left(e^{iax}\right)^2=0,$$
where $p_0,p_1,p_2\in\mathbb{C}(x)$, not all simultaneously constant, and $p_2$ not zero.
$$e^{iax}=-\frac{1}{2}\frac{p_1(x)\pm\sqrt{p_1(x)^2-4p_0(x)p_2(x)}}{p_2(x)}$$
$$-\frac{p_2(x)}{p_1(x)\pm\sqrt{p_1(x)^2-4p_0(x)p_2(x)}}e^{iax}=1$$
These equations are not in a form to be solved by elementary functions and Lambert W. See my answer at How to check if some equation can be solved using Lambert $\operatorname{W}$ function.
We need a generalization of Lambert W therefore.
If $-\frac{p_2(x)}{p_1(x)\pm\sqrt{p_1(x)^2-4p_0(x)p_2(x)}}$ is a rational expression of $x$, the generalized Lambert W of [Mezö/Baricz 2017] should be allowed as closed form.
[Abramowitz/Stegun 1970] Abramowitz, M.; Stegun, I.: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standard 1970
4.) Special functions and Elementary functions?
You could allow algebraic expressions of known Standard functions as closed form:
If the trigonometric functions can be decomposed into compositions of algebraic functions and other known algebraically independent Standard functions than $\exp$ and $\ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope such a generalization of Ritt's theorem could be proved someday for this class of functions. I made a suggestion at How to extend Ritt's theorem on elementary invertible bijective elementary functions?.
Best Answer
It is a consequence of Baker's Theorem, for example as listed here, that if the output of a trigonometric function is algebraic (and so rational numbers like $1/n$ count) then the input is either rational or transcendental. Since $\sqrt{n^2-1}$ is not rational but is algebraic for $n$ apart from 1, there are no other integer solutions to your equation.