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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?

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  • $\begingroup$ $\sin(\theta) = \sin \phi \implies \theta = \phi + 2 n \pi, n = 0, \pm 1, ...$ $\endgroup$ – jim Jun 28 '18 at 10:18
  • $\begingroup$ What is your assumption? What about $x=0?$ Remember that the functions are periodic, i.e. there are countable solutions. $\endgroup$ – gammatester Jun 28 '18 at 10:20
  • $\begingroup$ case II is peculiar. I would switch it for case 0: $4x=2x$. Further: what is the domain of $x$ here? $\endgroup$ – Vera Jun 28 '18 at 10:20
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    $\begingroup$ Would it not be simpler to use the fact that $\sin(4x)= \sin(2\cdot 2x) = 2\sin(2x)\cos(2x),$ so that $\sin(4x) = \sin(2x) \Leftrightarrow \sin(2x) \left(2\cos(2x) - 1\right) = 0$? $\endgroup$ – MSobak Jun 28 '18 at 10:23
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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: $$\left\{180^{\circ}k,30^{\circ}+60^{\circ}k|k\in\mathbb Z\right\}$$

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  • $\begingroup$ Why using degrees (I dislike them) if in the question radials are used? $\endgroup$ – Vera Jun 28 '18 at 13:08
  • $\begingroup$ @Vera I just like degrees. Here it's the same. If you want you can fixed it. $\endgroup$ – Michael Rozenberg Jun 28 '18 at 13:14
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    $\begingroup$ Thank you for explanation and giving me the liberty. I choose to respect your likes and will not fix. $\endgroup$ – Vera Jun 28 '18 at 13:18
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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.

$x=\frac{n\pi}{2},m\pi\pm\frac{\pi}{6}$

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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?"

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