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I've found two different ways to solve this trigonometric equation

$\begin{align*} \sin(2x)=\sin(x) \Leftrightarrow \\\\ 2\sin(x)\cos(x)=\sin(x)\Leftrightarrow \\\\ 2\sin(x)\cos(x)-\sin(x)=0 \Leftrightarrow\\\\ \sin(x) \left[2\cos(x)-1 \right]=0 \Leftrightarrow \\\\ \sin(x)=0 \vee \cos(x)=\frac{1}{2} \Leftrightarrow\\\\ x=k\pi \vee x=\frac{\pi}{3}+2k\pi \vee x=\frac{5\pi}{3}+2k\pi \space, \space k \in \mathbb{Z} \end{align*}$

The second way was:

$\begin{align*} \sin(2x)=\sin(x)\Leftrightarrow \\\\ 2x=x+2k\pi \vee 2x=\pi-x+2k\pi\Leftrightarrow \\\\ x=2k\pi \vee3x=\pi +2k\pi\Leftrightarrow \\\\x=2k\pi \vee x=\frac{\pi}{3}+\frac{2k\pi}{3} \space ,\space k\in \mathbb{Z} \end{align*}$

What is the correct one? Thanks

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Thanks, Alex, I was about to fix the whack-tex (forgotten \) errors. – ncmathsadist Jun 16 '12 at 23:14
Your solutions sets are actually the same, even though they are written differently. Notice that from $0$ to $2\pi$, the solutions are $x=0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi$, whichever answer you use, and in both cases, this pattern will repeat every $2 \pi$. – Jonas Kibelbek Jun 16 '12 at 23:16
up vote 7 down vote accepted

These answers are equivalent and both are correct. Placing angle $x$ on a unit circle, your first decomposition gives all angles at the far west and east sides, then all the angles $60$ degrees north of east, then all the angles $60$ degrees south of east.

Your second decomposition takes all angles at the far east side first. Then it takes all angles spaced one-third around the circle starting at 60 degrees north of east. You have the same solution set either way.

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Nice to see two ways to look at the problem – ncmathsadist Jun 16 '12 at 23:49

For what it's worth, the second one is "better" because it generalizes nicer. Imagine solving $\sin(3 x) = \sin(x)$ using the first method ($\sin(3 x) = 3\cos^2(x)\sin(x) - \sin^3(x)$). On the other hand, it's easy to see that $\sin(a x)= \sin(b x)$ will have an infinite number of solutions for any real $a$ and $b$ from the second method.

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You have all of the right ingredients.

The equation $\sin(x) = 0$ yields solutions $x = k\pi$ for $k\in \mathbb{Z}$.

The equation $\cos(x) = 0$ yields solutions $x = \pi/3, 5\pi/3$ in $[0,2\pi]$.

Since the cosine function is $2\pi$-periodic, you get the solutions

$$x = \pi(1/3 + 2k), \pi(5/3 + 2k), \qquad k\in\mathbb{Z}.$$

So your solution is the totality of all of these.

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