Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$ 
\sin 2x = \sin x;  \\ 0 \le x < 2 \pi $$
My method: 
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$  \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) = 0 $$
$$ \sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right) = 0 $$
Here one of the factors has to be $0$,
$$ \sin x = 0 \ \Rightarrow \  x = 0  \ or \ x = \pi $$
$$ \sin\left(\frac{x}{2}'\right) = 0 \ \Rightarrow \  x = 0  ;\  x \text{ can't be } \pi \text{ because of its range} $$
$$ \cos x = 0 \ \Rightarrow \  x = \frac{\pi}{2} \text{ or } \ x = \frac{3\pi}{2} $$
$$ \cos\left(\frac{3x}{2}\right) = 0 => x =  \frac{\pi}{3} \text{ or } x = \pi $$
So the solutions are : $$ 0, \pi,  \frac{\pi}{3} $$ 
I have seen other methods to solve this, so please don't post them. I'm really interested what's wrong with this one.
 A: $$\sin(2x)=\sin(x)\implies 2\cos(x)\sin(x)=\sin(x).$$
The elements of the form $k\pi$ are solutions (notice that if you divide the equation by $\sin(x)$ with not paying attention to this condition, you will loose solutions). Suppose $x\neq k\pi$. Then,
$$2\cos(x)\sin(x)=\sin(x)\implies 2\cos(x)=1\implies \cos(x)=\frac{1}{2}\implies x\in \left\{\pm\frac{\pi}{3}+2k\pi\mid k\in\mathbb Z\right\}.$$
Finally the solution are given by
$$\{k\pi\mid k\in\mathbb Z\}\cup\left\{\pm\frac{\pi}{3}+2k\pi\mid k\in\mathbb Z\right\}.$$
I let you take the solutions in $[0,2\pi[$.
A: There's nothing wrong up to the reduction to
$$
\sin\frac{x}{2}\cos\frac{3x}{2}=0
$$
Then you have either
$$
\sin\frac{x}{2}=0
$$
that is, $x/2=k\pi$ and $x=2k\pi$, or
$$
\cos\frac{3x}{2}=0
$$
so
$$
\frac{3x}{2}=\frac{\pi}{2}+k\pi
$$
and
$$
x=\frac{\pi}{3}+\frac{2k\pi}{3}
$$
Now let's examine the first set of solutions; you want to find the integers $k$ such that
$$
0\le 2k\pi<2\pi
$$
and this is only for $k=0$. For the second set of solutions,
$$
0\le\frac{\pi}{3}+\frac{2k\pi}{3}<2\pi
$$
becomes
$$
0\le 1+2k<6
$$
which gives $k\in\{0,1,2\}$.
Thus you find
$$
x\in\left\{0,\frac{\pi}{3},\pi,\frac{5\pi}{3}\right\}
$$
A: $\sin A=0\implies A=n\pi$
$\cos B=0\implies B=(2m+1)\pi/2$
$m,n$ are arbitrary integers
A: If we need to solve $$\sin \frac{x}{2} \cos \frac{3x}{2} = 0$$ for $0 \le x < 2\pi$, then the inequality condition is equivalent to $$0 \le x/2 < \pi,$$ and $$0 \le 3x/2 \le 3\pi.$$  Thus $\sin x/2 = 0$ admits only $x = 0$ in first interval; and in the second interval, we have solutions $$\frac{3x}{2} \in \left\{ \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \right\},$$ as all of these are between $0$ and $3\pi$.  Consequently, the second factor admits the solutions $$x \in \left\{\frac{\pi}{3}, \pi, \frac{5\pi}{3} \right\}.$$ The complete solution set is therefore $$x \in \left\{0, \frac{\pi}{3}, \pi, \frac{5\pi}{3} \right\}$$ in the desired interval.

A plot of the curves $$\color{blue}{y = \sin 2x}, \quad \color{orange}{y = \sin x}, \quad \color{green}{y = \sin 2x - \sin x}$$ is shown below:

As you can see, the green curve intersects the $x$-axis at the claimed values (as well as at $2\pi$ but this was excluded by the condition $0 \le x < 2\pi$).  You can also visually see that the $x$-coordinates at which the blue and orange curves coincide (corresponding to the equality $\sin 2x = \sin x$) are also the $x$-coordinates at which the green curve intersects the $x$-axis.
A: In restricting $0 \le x < 2\pi$ you are were inadvertently also restricting $0 \le 2x < 2\pi$ which is not a stated restriction.  
So $x = \pi$ is a solution as $\sin 2\pi$ does = $\sin \pi$ after all.
(Likewise $\sin 5\pi/3 = \sin 10\pi/3 = \sin 4\pi/3$).
So your restrictions are actually $0 \le 2x < 4\pi$ and for that matter $0 \le 3x/2 < 3\pi$.
So why did your answer not pick up $x = 5\pi/3$ as a potential answer.
As 
Then as $\cos(3x/2) = 0$ is potential answer, $0 \le 3x/2 < 3\pi$, so  $3x/2 = \pi/2, 3\pi/2, 5\pi/2$ so $x = \pi/3, \pi, 5\pi/3$.  
A: Other people have already pointed out we can rewrite like this:
$$2\sin(t)\cos(t) = \sin(t)$$
Now what most other people have not done, is this to substitute $\sin(t)$ and $\cos(t)$ for $y$ and $x$ and rewrite the problem to an algebraic variety:
$$\cases{2yx - y = 0\\
x^2+y^2=0}$$
This particular one is really silly easy to factor ( compared to systems of polynomial equations in general ):
$$\cases{y(2x-1) = 0\\x^2+y^2=0}$$
So either $2x-1 = 0$ or $y=0$
We recognize these equations as lines in the plane. The solutions are where they intercept the unit circle. Now it will be an easy exercise to substitute back and find all solutions.
