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Problem:

Find the general solution of $$\sin(mx)+\sin(nx)=0$$

My attempt: $$$$ $$\sin(mx)=-\sin(nx)$$ $$=\cos\left(\dfrac{\pi}{2}-mx\right)=\cos\left(\dfrac{\pi}{2}+nx\right)$$ Using $\cos\theta=\cos\alpha\Rightarrow \theta=2n\pi\pm \alpha,$$$$$ $$\text{CASE } 1:\theta=2n\pi+ \alpha$$ $$\dfrac{\pi}{2}-mx=2p\pi+\left(\dfrac{\pi}{2}+nx\right)$$ $$\Rightarrow x=\dfrac{-2p\pi}{m+n}$$$$$$ $$\text{CASE } 2:\theta=2n\pi- \alpha$$ $$\dfrac{\pi}{2}-mx=2q\pi-\left(\dfrac{\pi}{2}+nx\right)$$ $$\Rightarrow x=\dfrac{(2q-1)\pi}{n-m}$$$$$$ $$\Longrightarrow x=\dfrac{-2p\pi}{m+n} \text{ or } x=\dfrac{(2q-1)\pi}{n-m}$$

I checked my calculations again and again, but was unable to notice any flaw. However, the book I use categorically mentions the solutions for $x$ as $x=\dfrac{2j\pi}{m+n}$ or $x=\dfrac{(2k+1)\pi}{m-n}$ I would be truly grateful if somebody could please show me my errors. Many thanks in advance!

PS. Kindly note that $p,q,j,k\in \mathbb Z$

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  • $\begingroup$ Just take $p=-j$, $q=-k$ to get what the book says. $\endgroup$ – Samrat Mukhopadhyay Jul 21 '15 at 6:33
  • $\begingroup$ Possible duplicate. $\endgroup$ – user170039 Jul 21 '15 at 6:39
  • $\begingroup$ As long as you're not limiting the range of $x$, by saying for instance "find all solutions with $x\in [0,2\pi)$", and as long as all $x$ appear inside $\sin$, $\cos$, $\tan$ or similar functions, it is pretty hard to make an equation that doesn't have either zero or infinitely many solutions. You'd probably have to use periods that aren't rational multiples of one another. For instance, $\cos x +\cos(\pi x) =2$ has only one solution. $\endgroup$ – Arthur Jul 21 '15 at 6:44
  • $\begingroup$ @BetterWorld, multiple solutions do exist. As a concrete example take $m=2,n=1$ and take the solutions $x=2\pi/3,\ \pi$. $\endgroup$ – Samrat Mukhopadhyay Jul 21 '15 at 6:45
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Hint

Why not to use $$\sin(x)+\sin(y)=2\sin(\frac{x+y}2)\cos(\frac{x-y}2)$$ So, you just have a product to consider which will make life quite easier, I guess.

I am sure that you can take from here.

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  • $\begingroup$ @BetterWorld. You are very welcome ! $\endgroup$ – Claude Leibovici Jul 21 '15 at 6:40
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**HINT:**$$\sin mx=\sin(-nx)$$

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Notice, $$\sin(mx)+\sin(nx)=0$$ $$\implies \sin(mx)=-\sin(nx)$$ $$\implies \sin(mx)=\sin(-nx)$$ Now, writing the general solution of the equation $$mx=2k\pi+(-nx)$$ $$mx=2k\pi-nx$$$$(m+n)x=2k\pi$$$$\implies \color{blue}{x=\frac{2k\pi}{m+n}}$$ Or $$mx=(2k+1)\pi-(-nx)$$ $$mx-nx=(2k+1)\pi$$$$(m-n)x=(2k+1)\pi$$$$\implies \color{blue}{x=\frac{(2k+1)\pi}{m-n}}$$ Where, $\color{blue}{\text{k is any integer}}$

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