Finding integer solutions (by iterating over a variable) for an equation

Question

I have the following term

$$\sin \left(\arccos\left(\frac{300}{\cos (\arcsin (\frac{28}{x}))\cdot x+48}\right)\right) \cdot \left(\cos \left(\arcsin \left(\frac{28}{x}\right)\right)\cdot x + 48\right) $$

I want to figure out if there are any values of $x$ for which this term becomes an integer, or at least a value exact to a hundredth.

I already tried iterating the values in excel, which told me that for $x = 274$ the term becomes exactly $113$ However, excel apparently rounds the values as wolfram calculates a different result for $x = 274$. Furthermore, java and c give up and don't output any values at all (yes, I triple checked that I entered everything correctly). Besides that both apparently round the values as well (calculated the terms one after another for $x = 274$, got pretty much exactly $113$ as well).

Are there any other ways to solve this problem? It's important that the result is as exact as possible.

If anyone wonders, the term came up while solving a geometric problem (engineering).

Thank you!

Answer 1

Using $\sin(\arccos(x)) = \sqrt{1-x^2}$ and $\cos(\arcsin(x)) = \sqrt{1-x^2}$ :

• let $y = \cos \left(\arcsin \left(\frac{28}{x}\right)\right) \cdot x + 48 = x \cdot \sqrt{1 - \left(\frac{28}{x}\right)^2} + 48 = \sqrt{x^2 - 28^2} + 48$

• then $n = \sin\left(\arccos\left(\frac{300}{y}\right)\right) \cdot y = y \cdot \sqrt{1 - \left(\frac{300}{y}\right)^2} = \sqrt{y^2 - 300^2}$

For $x=274$ the above calculate to $y=48 + 2 \sqrt{18573}$ and $n=2 \sqrt{48 \sqrt{18573} - 3351} \approx112.97$ so the expression is not an integer.

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[ EDIT ]  To answer the main question about solutions in $x$ such that the expression $n$ is an integer, one can choose any integer $n \in \mathbb{N}$, then calculate $y=\sqrt{n^2+300^2}$ and $x=\sqrt{(y-48)^2+28^2}$.

For example, $n=0$ $\implies$ $y=300$ $\implies$ $x=\sqrt{(300-48)^2+28^2}=\sqrt{64288}=28\sqrt{82}$.

A finite number of values $x$ will yield integer values for $y$, the smallest one being $x=\sqrt{66833}$ which gives $y=305$ and $n=55$. Those can be found by enumerating the pythagorean triples having $300$ as a side, and the hypothenuse given by $y^2 = 300^2+n^2$.

Answer 2

Using the following easily verified identities the expression can be rewritten as an algebraic expression.

$$ t\cdot\sin\left(\arccos\left(\frac{c}{t}\right)\right)=t\cdot\cos\left(\arcsin\left(\frac{c}{t}\right)\right) =\sqrt{t^2-c^2}$$

Then

$$\cos \left(\arcsin \left(\frac{28}{x}\right)\right)\cdot x=\sqrt{x^2-784}$$

and

\begin{align} & \sin \left(\arccos\left(\frac{300}{\sqrt{x^2-784}+48}\right)\right) \cdot \left(\sqrt{x^2-784} + 48\right)\\ &=\sqrt{(\sqrt{x^2-784}+48)^2-90000} \end{align}

In general

$$ \sin \left(\arccos\left(\frac{A}{\cos (\arcsin (\frac{B}{x}))\cdot x+C}\right)\right) \cdot \left(\cos \left(\arcsin \left(\frac{B}{x}\right)\right)\cdot x + C\right)$$

simplifies to the algebraic expression

$$ \sqrt{(\sqrt{x^2-B^2}+C)^2-A^2} $$