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Is there a theorem stating that a general formula for the solution to the equation \begin{equation} \sin(ax)+\sin(bx)=0 \end{equation} does not exist in terms of elementary functions?

I don't know what keywords to search for to better understand this problem; on google I keep finding methods to find the numerical solution rather than an algebraic discussion.

When $a,b$ are integers, is this problem related to Galois theory, since in this case $\sin(ax)$ and $\sin(bx)$ can be expressed as polynomials in $\sin(x)$ and $\cos(x)$?

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  • $\begingroup$ Actually a formula exists, and it is quite simple. You can use the fact that $\sin A = \sin B$ if and only if $A = n \pi +(-1)^n B$ $\endgroup$ – Crostul Jan 8 '17 at 10:34
  • $\begingroup$ Yes, I just realized that... $\endgroup$ – marco trevi Jan 8 '17 at 10:34
  • $\begingroup$ I think this is unrelated to Galoia theory. $\endgroup$ – Takahiro Waki Jan 8 '17 at 12:17
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$$\sin(ax)=-\sin(bx)=\sin(-bx)$$

$$ax=n\pi+(-1)^n(-bx)$$ where $n$ is any integer

If $n$ is odd$=2m+1$(say), $ax=(2m+1)\pi+bx$

If $a\ne b, x=\dfrac{(2m+1)\pi}{a-b}$

What if $a=b?$

What if $n$ is even $=2m$(say)?

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  • $\begingroup$ I just realized the same thing. I am really trying to figure out more complicated equations like a generic sum of sines and cosines with different frequencies. I think I underthought the wording of the question. My fault. $\endgroup$ – marco trevi Jan 8 '17 at 10:36
  • $\begingroup$ what is in the case $$a=b$$? $\endgroup$ – Dr. Sonnhard Graubner Jan 8 '17 at 10:37
  • $\begingroup$ @Dr.SonnhardGraubner, Updated the answer for better understanding $\endgroup$ – lab bhattacharjee Jan 8 '17 at 10:39
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Hint. One may recall that $$ \sin(ax)+\sin(bx)=2\cdot\sin\left(\frac{ax+bx}2\right) \cdot \sin\left(\frac{ax-bx}2\right). $$

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