How to solve $\sin 5x = - \sin 3x?$

Hint: This problem deals with solving an equation which is solved by the help of trigonometric sum to product identities. Given an equation, we have to find the number of solutions of the equation. Which means that there are several solutions of the equation and hence we have to find the general solution of the equation. The formula used here is the trigonometric sum to product formula, which is given by:
$ \Rightarrow \sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$
If $\sin x = 0$, then $x$ would be a multiple of $\pi $.
If $\cos x = 0$, then $x$ would be an odd multiple of $\dfrac{\pi }{2}$.

Complete step-by-step solution:
Given a trigonometric equation in terms of trigonometric ratios of sine angles, which is given by:$ \Rightarrow \sin 5x = - \sin 3x$
Now consider the given trigonometric equation as given below:
$ \Rightarrow \sin 5x = - \sin 3x$
Rearranging the above equation as shown below:
$ \Rightarrow \sin 5x + \sin 3x = 0$
The above equation is in the form of standard trigonometric sum to product identities. Hence applying the trigonometric sum to product identity to the above obtained equation, as given below:
$ \Rightarrow \sin 5x + \sin 3x = 0$
$ \Rightarrow 2\sin \left( {\dfrac{{5x + 3x}}{2}} \right)\cos \left( {\dfrac{{5x - 3x}}{2}} \right) = 0$
$ \Rightarrow 2\sin \left( {\dfrac{{8x}}{2}} \right)\cos \left( {\dfrac{{2x}}{2}} \right) = 0$
$ \Rightarrow 2\sin 4x\cos x = 0$
$ \Rightarrow \sin 4x\cos x = 0$
Here either $\sin 4x = 0$ or $\cos x = 0$, so considering both the cases, as shown below:
First consider $\sin 4x = 0$, as given below:
$ \Rightarrow \sin 4x = 0$
$ \Rightarrow 4x = n\pi $
$ \Rightarrow x = \dfrac{{n\pi }}{4}$
Here $n$ is an integer, where $n = 0,1,2,....$
Now consider $\cos x = 0$, as given below:
$ \Rightarrow \cos x = 0$
$ \Rightarrow x = \left( {2n + 1} \right)\dfrac{\pi }{2}$
Here $n$ is an integer, where $n = 0,1,2,....$
Hence the solution of $x$ is both the union of the considered cases.
$ \Rightarrow x = \dfrac{{n\pi }}{4}$ and $x = \left( {2n + 1} \right)\dfrac{\pi }{2}$
$\therefore x = \left\{ {\dfrac{{n\pi }}{4}} \right\} \cup \left\{ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right\};n \in 0,1,2,...$

The solution of the equation $\sin 5x = - \sin 3x$ is $x = \left\{ {\dfrac{{n\pi }}{4}} \right\} \cup \left\{ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right\};n \in 0,1,2,...$

Note: While solving this problem, we need to understand that throughout the problem we used one of the trigonometric sum to product formula, there are totally four such trigonometric sum to product formulas, which are given by:
\[ \Rightarrow \sin C + \sin D = 2\sin (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})\]
\[ \Rightarrow \sin C - \sin D = 2\cos (\dfrac{{C + D}}{2})\sin (\dfrac{{C - D}}{2})\]
\[ \Rightarrow \cos C + \cos D = 2\cos (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})\]
\[ \Rightarrow \cos C - \cos D = - 2\sin (\dfrac{{C + D}}{2})\sin (\dfrac{{C - D}}{2})\]