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.
$\endgroup$ 13 Answers
$\begingroup$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 $$x=\frac{7\pi}{18}+\frac{2k\pi}3$$
$\endgroup$ $\begingroup$Hint: check that we have
$$\sin\alpha=\sin\beta\iff\begin{cases}\alpha=\beta\;,\;\;or\\{}\\\alpha=\pi-\beta\end{cases}\;\;\;+2k\pi\;,\;\;k\in\Bbb Z$$
$\endgroup$ $\begingroup$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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