TITLE: Solving trigonometry equation - getting variables out from the sin functions
QUESTION [0 upvotes]: I want to solve the following equation here in terms of $w$ ($w = \dots$).
$$\sin(x) + \sin(2^{\frac{n+12}{12}}x) = \sin(x+w) + \sin(2^{\frac{n+12}{12}}(x+w))$$

REPLY [0 votes]: I'll assume that we're looking for a $w$ that makes the expressions equal for all, rather than for some particular $x$. Using, e.g., Fourier series, one can show that unless $2^{(n + 12) / 12} = 1$, that is, $n = -12$, we must have
\begin{align}
\sin x &= \sin (x + w) \\
\sin 2^{(n + 12) / 12} x &= \sin 2^{(n + 12) / 12} (x + w)
\end{align}
So, $w$ is a solution iff both $w$ and $2^{(n + 12) / 12} w$ are multiples of $2 \pi$. In particular this implies that $2^{(n + 12) / 12}$ is rational, which forces the exponent to be integral, that is, that $n$ be a multiple of $12$.
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