# Why is there no general solution to $\sin(ax)+\sin(bx)=0$?

Is there a theorem stating that a general formula for the solution to the equation $$\sin(ax)+\sin(bx)=0$$ 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 $\cos(x)$?

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

$$\sin(ax)=-\sin(bx)=\sin(-bx)$$

$$ax=n\pi+(-1)^n(-bx)$$ 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)?

• 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. – marco trevi Jan 8 '17 at 10:36
• what is in the case $$a=b$$? – Dr. Sonnhard Graubner Jan 8 '17 at 10:37
• @Dr.SonnhardGraubner, Updated the answer for better understanding – lab bhattacharjee Jan 8 '17 at 10:39

Hint. One may recall that $$\sin(ax)+\sin(bx)=2\cdot\sin\left(\frac{ax+bx}2\right) \cdot \sin\left(\frac{ax-bx}2\right).$$