# 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.

$$\cos(2x)[1-2\sin(2x)]=0$$

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?

• If it's legit to divide by $\cos(2x)$ then it must be the case that $\cos(2x) \neq 0$. – user4894 Mar 29 '15 at 0:35
• That is correct. It will give you two cases: if $\cos(2x)\neq 0$... solve as you did (with division). If $\cos(2x)=0$... solve that equation for the remaining solutions. – TravisJ Mar 29 '15 at 0:35
• By dividing both sides by $\cos(2x)$ you are pre-supposing that the solution $\cos(2x)=0$ is not valid as otherwise you cannot divide both sides by zero. – Mufasa Mar 29 '15 at 0:35
• @user4894 Ah yes, that makes sense. But how would I have known that case is true, starting at $4\sin(2x)\cdot\cos(2x)=2\cos(2x)$? – user1534664 Mar 29 '15 at 0:41
• you can divide by $2\cos 2x$ without any trouble for most values of $x.$ sometimes the exceptional ones are the critical ones. i am thinking of the zeros of $f'$ – abel Mar 29 '15 at 1:28

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.

• Thank you, sir. Will (2) always be applicable? – user1534664 Mar 29 '15 at 0:52
• @user1534664 No, so maybe you should practice both methods. – GFauxPas Mar 29 '15 at 1:14

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.)