Solve: $\sin 2x + \sin 4x = 2 \sin 3x$
$\sin 2x + \sin 4x = 2 \sin 3x$
i.e. $\;$ $2 \sin \left(\dfrac{2x + 4x}{2}\right) \cos \left(\dfrac{4x - 2x}{2}\right) = 2 \sin 3x$
i.e. $\;$ $\sin 3x \; \cos x = \sin 3x$
i.e. $\;$ $\sin 3x \left(\cos x - 1\right) = 0$
i.e. $\;$ $\sin 3x = 0$ $\;$ or $\;$ $\cos x - 1 = 0$
Now, $\;$ $\sin 3x = 0$ $\implies$ $3x = n \pi$ $\;\;$ i.e. $\;$ $x = \dfrac{n \pi}{3}, \;\;\; n \in Z$
Principal value of $\cos x$ lies in $\left[0, \pi\right]$
Now, $\cos x - 1 = 0$ $\implies$ $\cos x = 1 = \cos 0$ $\;\;\;$ $\left(\because \;\cos x = +ve\right)$
$\implies$ $x = 2 m \pi, \;\;\; m \in Z$