Given trigonometric equation
$$\sin(4x) = \sin(2x)$$
I'm trying to obtain $x$.
Case I)
$$\sin(\pi - 4x) = \sin(2x)$$
$$\pi - 4x = 2x$$
$$\boxed {\dfrac{\pi}{6} = x}$$
Case II)
$$\sin(2\pi - 4x) = \sin(2x)$$
$$2\pi - 4x = 2x$$
$$\boxed {\dfrac{\pi}{3} = x}$$
Is my assumption correct?
On
Your second case is clearly not true, for if $x=\frac{\pi}{3}$
$\sin(4x)=-\frac{\sqrt3}{2}$
and
$\sin(2x)=\frac{\sqrt3}{2}$
so this counter-example disproves your conjecture ... for all you actually asked was "Is my assumption correct?"
On
Change the form of $\sin(4x)=\sin(2x)$ to get $2\sin(2x)\cos(2x)=\sin(2x)$
If $\sin(2x)=0$, $2x=n \pi$ and $x=\frac{n \pi}{2}.$ Where $n$ is an integer.
If $\sin(2x) \neq 0,$ $\cos(2x)=\frac{1}{2}$ $\space$ $2x=2m\pi\pm\frac{\pi}{3}$ $\space$ $x=m\pi\pm\frac{\pi}{6}.$
Where $m$ is an integer.
$x=\frac{n\pi}{2},m\pi\pm\frac{\pi}{6}$
No! $60^{\circ}$ is not a root.
Also, you equation has infinitely many solutions. $$4x=2x+360^{\circ}k,$$ where $k$ is an integer number, which is $$x=180^{\circ}k$$ or $$4x=180^{\circ}-2x+360^{\circ}k,$$ which gives $$x=30^{\circ}+60^{\circ}k$$ and we got the answer: $$\left\{180^{\circ}k,30^{\circ}+60^{\circ}k|k\in\mathbb Z\right\}$$