Trigonometry Equations (Sum to Product)

Question

I am struggling with how to determine which solutions are not part of the solution set. For instance, in $\sin(6x)+\sin(x)=0$, I used the sum to product theorem and got $\sin(7x)\cos(5x)=0$. I got solutions $(\pi/7)n$ but I didn't discard some of the solutions because they wouldn't work for the $\sin(6x)+\sin(x)=0$. Do you know how to solve this by using a simpler method?

Answer 1

An alternative method: $$\eqalign{\sin6x+\sin x=0\quad &\Leftrightarrow\quad \sin x=-\sin6x\cr &\Leftrightarrow\quad x=-6x+2k\pi\ \hbox{or}\ x=6x+(2k+1)\pi\cr &\Leftrightarrow\quad x=\frac{2k\pi}7\ \hbox{or}\ x=-\frac{(2k+1)\pi}{5}\ .\cr}$$

Answer 2

An alternative way of solving: $$\sin 6x + \sin x = 0$$ $$\implies\sin 6x = -\sin x$$ $$\implies\sin 6x = sin(-x)$$ $$\implies 6x = n\pi + (-1)^{n+1}x $$ $$\implies 6x-(-1)^{n+1}x= n\pi$$ $$\implies x(6+(-1)^n)= n\pi$$ $$\implies x = \frac{n\pi}{6+(-1)^n}$$, where n $\in \Bbb{ Z}$.