General Solution of $\sin(mx)+\sin(nx)=0$ 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 $p,q,j,k\in \mathbb Z$

**HINT:**$$\sin mx=\sin(-nx)$$
