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
$\endgroup$ 23 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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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