There is an equation
$$\sin2\theta=\sin\theta$$
We need to show when the right-hand side is equal to the left-hand side for $[0,2\pi]$.
Let's rewrite it as
$$2\sin\theta\cos\theta=\sin\theta$$
Let's divide both sides by $\sin\theta$ (then $\sin\theta \neq 0 \leftrightarrow \theta \notin \{0,\pi,2\pi\}$)
$$2\cos\theta=1$$
$$cos\theta=\frac{1}{2}$$
$$\theta\in\left\{\frac{\pi}{3},\frac{5\pi}{3}\right\}$$
Now, let's try something different.
$$2\sin\theta\cos\theta=\sin\theta$$
$$2\sin\theta\cos\theta-\sin\theta=0$$
$$\sin\theta(2\cos\theta-1)=0$$
We can have the solution when $\sin\theta=0$.
$$\sin\theta=0$$
$$\theta \in \left\{0,\pi,2\pi\right\}$$
And when $2\cos\theta-1=0$.
$$2\cos\theta-1=0$$
$$2\cos\theta=1$$
$$\cos\theta=\frac{1}{2}$$
$$\theta \in \left\{\frac{\pi}{3},\frac{5\pi}{3}\right\}$$
Therefore the whole solution set is
$$\theta \in \left\{0,\pi,2\pi,\frac{\pi}{3},\frac{5\pi}{3}\right\}$$
This is the correct solution.
Why is this happening? In the first approach, the extra solution given by $\sin\theta=0$ is not only non-appearing but actually banned. Both approaches look valid to me, yet the first one yields less solutions than the second one. Is the first approach invalid in some cases? This is not the only case when this happens, so I'd like to know when I need to use the second approach to solve the equation, so I don't miss any possible solutions.
Best Answer
Whenever you divide both sides of an equation by something, you are assuming that the thing you're dividing by is nonzero, because dividing by $0$ is not valid.
So going from $2 \sin \theta \cos \theta = \sin \theta$ to $2 \cos\theta = 1$ is only valid when $\sin\theta \ne 0$.
In general, all this means is that you need to check the $\sin \theta = 0$ case separately. For example, going from $2 \sin\theta \cos\theta = 1$ to $2 \cos\theta = \frac1{\sin\theta}$ is also only valid when $\sin\theta \ne 0$, but you don't lose any solutions by doing this, because values of $\theta$ for which $\sin\theta=0$ weren't solutions to begin with.
But in this particular case, when $\sin\theta = 0$, the equation $2\sin\theta \cos\theta = \sin\theta$ is satisfied, so it's correct to go from this to $$ 2\cos\theta = 1 \text{ or } \sin\theta = 0. $$