From Förberedande kurs i matematik 1

If we consider for a moment the equality

\displaystyle \sin u=\sin v\quad \quad \quad (*)

where \displaystyle u has a fixed value, there are usually two angles \displaystyle v in the unit circle which ensure that the equality holds:

\displaystyle v=u and \displaystyle v=\pi -u

(The only exception is when \displaystyle u={\pi }/{2}\;\text{ } or \displaystyle u=3{\pi }/{2}\;\text{ }, in which case \displaystyle u and \displaystyle \pi -u\text{ } correspond to the same direction and there is only one angle \displaystyle v which satisfies the equality.)

We obtain all the angles \displaystyle v which satisfy (*) by adding multiples of \displaystyle \text{2}\pi ,

\displaystyle v=u+2n\pi and \displaystyle v=\pi -u+2n\pi

where \displaystyle n\text{ } is an arbitrary integer.

If we now go back to our equation

\displaystyle \sin 3x=\sin x

the reasoning above shows that the equation is only satisfied when

\displaystyle 3x=x+2n\pi or \displaystyle 3x=\pi -x+2n\pi

If we make \displaystyle x the subject of each equation, we obtain the full solution to the equation:

\displaystyle \left\{ \begin{array}{*{35}l} x=0+n\pi \\ x=\frac{\pi }{4}+\frac{1}{2}n\pi \\ \end{array} \right.