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I got a this equation to solve and $\sin(2x) = \sin(x-\frac{\pi}{6})$

They was nice to give me for possible answers it could be.

x = $\frac{\pi}{6}$

x = $\frac{\pi}{18}$

x = $\frac{7\pi}{18}$

x = $\frac{55\pi}{18}$

After i calculated i thought it was x = $\frac{7\pi}{18}$ because I first solved $\sin(2x) = \sin(x-\frac{\pi}{6})$ and got it to around 1.22 and x = $\frac{7\pi}{18}$ was the only option that got the same value. But it turned out to be wrong.

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  • $\begingroup$ Please use \sin instead of sin in function names, compare the difference $\sin$ vs $sin$ $\endgroup$ – Alice Ryhl Sep 2 '14 at 19:54
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Notice $\sin A=\sin B$ iff $A=B+2k\pi$ or $A=\pi-B+2k\pi$, $k\in\Bbb Z$.

So, $\sin 2x=\sin (x-\frac{\pi}6)$ iff

$$2x=x-\frac{\pi}6+2k\pi$$ or $$2x=\pi-x+\frac{\pi}6+2k\pi$$

That is,

$$x=-\frac{\pi}6+2k\pi$$ or $$x=\frac{7\pi}{18}+\frac{2k\pi}3$$

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Hint: check that we have

$$\sin\alpha=\sin\beta\iff\begin{cases}\alpha=\beta\;,\;\;or\\{}\\\alpha=\pi-\beta\end{cases}\;\;\;+2k\pi\;,\;\;k\in\Bbb Z$$

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Taking the $\arcsin$, both members are equal or complementary, to an even multiple of $\pi$. $$2x=x-\frac\pi6+2k\pi\lor\pi-2x=x-\frac\pi6+2k\pi$$ $$x=-\frac\pi6+2k\pi\lor x=\frac{7\pi}{18}+\frac23k\pi$$ Possible answers are

$$-\frac\pi6,\frac{11\pi}6,\frac{23\pi}6,\frac{35\pi}6..., \color{blue}{\frac{7\pi}{18}},\frac{19\pi}{18},\frac{31\pi}{18},\frac{43\pi}{18},\color{blue}{\frac{55\pi}{18}},\frac{67\pi}{18}...$$

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