$\sin \theta = \frac{4}{5}$
Does an $n \in \mathbb{Z_+}$ exists which satisfies the equation $n = \frac{2k\pi}{\theta}$ where $k \in \mathbb{Z}$ ?
I tried to write $\theta$ as $\theta = x\pi$ and after that $n = \frac{2k}{x}$. But the problem is that I am not sure if $x \in \frac{\mathbb{R}}{\mathbb{Q}}$ or $x\equiv 0\pmod 2$ so I can't determine if $n \in \mathbb{Z_+}$
This is a question about (non-trivial) number theory, not really about trigonometry.
If $\sin\theta=\frac45$, then $\cos\theta=\pm\frac 35$. The question whether $n\theta$ is a multiple of $2\pi$ is equivalent to asking wheter $(\pm\frac35+i\frac45)^n=1$ or equivalently $(\pm3+4i)^n=5^n$ for some integer $n\ge 1$.
In the ring $\Bbb Z[i]$, we have unique prime factorization, but not all our usual primes are primes in this ring. In particular, $5$ is the product of two primes $5=p_1p_2$, namely of $p_1=4+3i$ and $p_2=4-3i$. Note that primes are considered equivalent in $\Bbb Z[i]$ if they differ only by a unit factor ($1$, $-1$, $i$, or $-i$) and that in this sense $p_1$ and $p_2$ are not equivalent. We conclude that the unique prime factorization of $5^n$ is $p_1^np_2^n$. On the other hand, $3+4i=ip_2$ and hence $(3+4i)^n=i^np_2^n$ and similarly, $(-3+4i)^n=i^np_1^n$. At any rate, the prime factorizations of $5^n$ and $8\pm3+4i)^n$ differ for $n>0$. Therefore $(\pm3+4i)^n\ne 5^n$.