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The equation is: $\sin x = \sin 2x$

I recognize that $\sin 2x$ is a double angle and use the double angle identity, so it becomes:

$$\begin{array}{rrcl} &\sin x &=& 2 \sin x \cos x \\ \implies& \sin x - 2 \sin x \cos x &=& 0 \end{array}$$

Then I am stuck... Not sure how to proceed.

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    $\begingroup$ Hint: $\sin x (1 - 2 \cos x)=0$ means (at least) one of the factors must be $0\,$. $\endgroup$
    – dxiv
    May 15, 2017 at 1:11
  • $\begingroup$ Use the zero product property: If $ab=0$ and $a,b \in \mathbb{R}$ then either $a=0$ or $b=0.$ $\endgroup$ May 15, 2017 at 1:15

3 Answers 3

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\begin{align} \sin x & = \sin 2x \\ &= 2 \sin x \cos x \end{align} Then \begin{align} \sin x - 2 \sin x \cos x = 0 \end{align} So $$\sin x (1 - 2\cos x ) = 0$$ This imples that $\sin x = 0$ or $(1-2\cos x) = 0$. $\sin x = 0$ implies that $x = n \pi$. Also $(1 - 2\cos x ) = 0$ implies that $\cos x = \frac{1}{2}$ and so $x = 2n \pi + \frac{5\pi}{3}, 2n \pi + \frac{\pi}{3}$. So that the solution set is $$\left \{n \pi,2n \pi + \frac{5\pi}{3}, 2n \pi + \frac{\pi}{3}: n \in\mathbb{Z}\right\}$$

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Hint:

In general if $\sin x=\sin y,$

$x=n\pi+(-1)^ny$ where $n$ is any integer.

Check for $n$ is even and $n$ is odd one by one

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$$\sin x = \sin 2x \implies \sin x = 2\sin x \cos x \implies \sin x - 2\sin x \cos x = 0 \implies \sin x (1-2\cos x) = 0$$

That means $\sin x = 0$ or $\cos x = \frac{1}{2}$

$\sin x = 0$ is true when $x = k\pi$

$\cos x = \frac{1}{2}$ is true when $x = \frac{\pm (1+ 2k\pi)}{3}$ or $x = \frac{\pm (5+ 2k\pi)}{3}$

So this statement holds when $x = k\pi$ or $x = \frac{\pm (1+ 2k\pi)}{3}$ or $x = \frac{\pm (5+ 2k\pi)}{3}$.

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    $\begingroup$ Be careful. $\cos(2\pi/3) = -1/2$. $\endgroup$ May 15, 2017 at 1:18
  • $\begingroup$ $\cos(2\pi/3)$ isn't even used here $\cos(-\pi/3) = \cos(5\pi/3)$ $\endgroup$
    – Eric Lee
    May 15, 2017 at 1:20
  • $\begingroup$ You said that $\cos x = 1/2$ is true when $x = \pm k\pi/3$, where $k$ is an integer. $\endgroup$ May 15, 2017 at 1:21
  • $\begingroup$ Sorry Think it's fixed? $\endgroup$
    – Eric Lee
    May 15, 2017 at 1:25
  • $\begingroup$ No. Observe that when $k = 0$, your answer is $x = \pm 1/3$ or $x = \pm 5/3$. You could write $x = \pm \frac{\pi}{3} + 2k\pi, k \in \mathbb{Z}$. $\endgroup$ May 15, 2017 at 1:32

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