Why cant I solve the equation $4\sin(2x)\cdot \cos(2x)=2\cos(2x)$ by dividing both sides by $\cos(2x)$?? This assignment assumes the domain is $[0, \pi]$.
Why can't I solve the following equation by dividing both sides by $\cos(2x)$?
$$4\sin(2x)\cdot \cos(2x)=2\cos(2x)$$
If I would continue to do so, wouldn't $\cos(2x)$ cancel out on both sides, leaving me with
$$4\sin(2x)=2$$
Which will then be solvable.
According to khanacademy's answer it seems like I would now miss out on two solutions.
Khanacademy's answer:
$$\cos(2x)[1-2\sin(2x)]=0$$
Then either 
$$\cos(2x)=0\\ \sin(2x)=\frac{1}{2}$$
As we can see I don't have $\cos(2x) = 0$ "set up". But what I did was just a simple division, which seems like valid algebra, so how can I tell which approach to use in other scenario's?
 A: Consider the equation:
$$xy = x$$
You can observe that $x=0$ satisfies this equation. But if you were to divide both sides by $x$ you would get
$$y = 1$$
which is one solution, but you lost the other solution! The problem is that you can't divide by zero, so if you divide both sides by $x$, you're taking for granted that $x$ isn't zero.
Two ways around this:
$(1)$ Check if the thing you're divididing by can be equal to zero. If it can be, separate the problem into cases: case $x=0$, case $x \ne 0$.
$(2)$ Don't divide both sides by the questionable factor. For example, consider $xy - x = x(y-1) = 0$ in our example. If a product is zero one of the factors must be zero.
A: Any time you divide by something that is not a non-zero constant 
(such as $3$, $2\pi$, or $\sqrt{5+\sqrt{2}}$),
you have to ask yourself whether it's possible that you're dividing by zero.
There are then two ways you can deal with that question.
One way is you can set about proving that the quantity can never be zero.
Sometimes that makes sense to do (for example, if $x$ is real you can always
divide by $x^2 + 1$). If the domain of $x$ were limited to $[0,\frac12]$
then it would be the case that $\cos(2x)\neq 0$ and you could prove it.
But sometimes, as in this case, it is simply not true that the thing
you want to divide by is always non-zero.
The best you might accomplish by trying to prove it is never zero is
that you might actually find the zeros and see that they are useful
in reaching a solution.
Another way is to consider the possibility that the quantity might
sometimes be zero, and split your solution into two cases.
In one case, you assume the quantity is non-zero, divide by it,
and proceed from there.
But at some point (either before or after working out the non-zero case)
you must look at the case where the quantity is zero.
One way to keep track of this is that from the equation
$$4\sin(2x)\cdot \cos(2x)=2\cos(2x)$$
you determine that this is equivalent to
$$4\sin(2x) = 1  \qquad \mbox{OR} \qquad \cos(2x) = 0.$$
You work the formula on the left of the "or", finding zero, one, or more
solutions, and you work the formula on the right side, finding
zero, one or more solutions.
Since the original formula is satisfied if either of the two new formulas
is satisfied, you get your complete set of solutions by taking
the union of the two solution sets of the new formulas.
(If one of those formulas has no solution, its solution set is empty and
it adds no solutions to the final solution set. You have in effect then
proved that it was OK to divide by this quantity in the first place.)
