Question: Solve $\displaystyle \displaystyle sin(5x)+sin(x)=0$ for $\displaystyle 0\leq x < \pi$.

I know ways to solve this by expanding sin(5x) but I'm wondering if the method I'm attempting below is correct and how to proceed with it:

$\displaystyle \displaystyle \sin 5x = -sin x \implies \sin 5x = \sin {-x}$

$\displaystyle \implies \sin 5(x-2k_1 \pi) = \sin -(x-2k_2 \pi)$

where $\displaystyle k_1$ and $\displaystyle k_2$ are integers.

$\displaystyle \displaystyle \implies 5x-10k_1 \pi = -x+2k_2 \pi \implies x=\frac{\pi(2k_2+10k_1)}{6} = \frac{\pi(k_2+5k_1)}{3}$

Now the solution interval is between 0 and pi so we have the following inequality:

$\displaystyle \displaystyle 0\leq \frac{k_2+5k_1}{3}<1 \implies 0\leq k_2+5k_1<3$

There are infinite integer solutions to this inequality so how do I limit the solutions further?