Solving $\sin(2v) = \sin(v)$ $$\sin(2v) = \sin(v)$$
Why can't this equation be solved by setting:
$$2v = v + 2\pi n \quad \leftrightarrow  \quad v = 2\pi n\\2v = \pi - v + 2\pi n \quad\leftrightarrow \quad 3v = \pi + 2\pi n  \quad \leftrightarrow \quad v = \frac{\pi}{3} + \frac{2\pi n}{3}$$
This is how I've seen other similar equations being solved.
Please note that I was able to solve the equation by setting $\sin(2v) = 2\sin(v)\cos(v)$, I just want to know why this particular equation not can be solved the way I presented.
The correct solutions are (according to my literature): $v = \pi n$ or $v = \pm \frac{\pi}{3} + 2\pi n$
 A: You can solve it by your method as well and your answer is right.

EDIT
Let us look at the solutions of the form $\dfrac{\pi}3 + \dfrac{2\pi n}3$. $n$ an be only of the form $3k$ or $3k \pm 1$.


*

*If $n=3k$, we obtain the solution to be $2k\pi + \dfrac{\pi}3$

*If $n=3k+1$, we obtain the solution to be of the form $2k\pi + \pi$

*If $n=3k-1$, we obtain the solution to be of the form $2k\pi - \dfrac{\pi}3$


The above along with the solution $2k \pi$ gives you all the possibilities as the other solution.
A: Note that both your solutions and the solutions in the literature repeat with a period of $2\pi$, so it's enough to check that they're the same within the $2\pi$-long interval $[-\pi, \pi)$.
In that interval, the literature's solutions are:


*

*$v=\pi n$, leading to $v=-\pi$ or $v=0$ (for $n=-1$ or $n=0$)

*$v=\pm \frac{\pi}{3} + 2\pi n$, leading to $v = -\frac{\pi}{3}$ or $v=\frac{\pi}{3}$ (for $n=0$).


while your solutions are:


*

*$v=2\pi n$, leading to $v=0$

*$v=\frac{\pi}{3}+\frac{2\pi n}{3}$, leading to $v=-\pi$, $v=-\frac{\pi}{3}$, or $v=\frac{\pi}{3}$ (for $n=-2,-1,0$ respectively).


That is, you get all the same solutions as the book; they're just split up differently.
