Is it legal to divide with a variable

Question

I know that I can't divide by $\cos x$ when solving an equation because it might be equal to $0$. In a worksheet, my teacher divided by $\cos x$. I want to know when can I divide by it.

$$\frac{8 \sin^2(x) (2\cos^3x-\cos x)}{\cos(x)} = \frac{4\cos^5x+\cos x-4\cos^3x}{\cos(x)}$$

Answer 1

You can divide by anything that is not constantly $0$ as long as you treat separately the case when that thing is $0$. That way, you obtain some solutions for $\cos{x}=0$ and some other solutions for $\cos{x}\neq 0$.

Answer 2

We can divide for $\cos x$ but we have to consider the extra condition $\cos x\neq 0$. We can then solve the new expression and find solutions from which we need to exclude that for which $\cos x =0$. Note that finally the solution for $\cos x=0$ must be checked on the original expression.

Note also that the same method applies for any expression $g(x)$ and if $g(x)\neq 0$ we don’t need extra conditions.

Answer 3

Here are two example that generalize easily.

1) Say I want to solve the equation $$x^2 = 2x.$$ Dividing by the variable $x$ yields $$x = 2.$$ But of course, $0^2 = 2(0)$. Because I divided by $x$ without first investigating what happens when $x=0$ I lost the data that 0 is a solution. The correct way to proceed if I really want to divide by $x$ would be to assert that I am considering $x \neq 0$ before dividing, and then consider the case $x = 0$ separately.

In light of the comment by vrugtehagel, I will include another example that may seem different. In fact, I teach pre calculus at a college and it does tend to confuse my students.

2) Suppose we want to solve $$3x = 2x.$$ I see right away that dividing by the variable might be a good idea, but of course, we cannot divide by 0, so first I inspect if 0 is or is not a solution. Evaluating at $x = 0$, I get 3(0) = 2(0), thus $0=0$ which is true so $x=0$ is a solution. Now I consider the same equation $3x = 2x$, but for $x \neq 0$. Since $x \neq 0$ it is permissible to divide by $x$, yielding $3 = 2$. This is the step where my pre-calc students tend to panic, because 3 is obviously not equal to 2, what does this mean!? There is no need to panic, this is still helping us solve our equation. One simply needs to know that the logical connective between elementary and legal algebraic manipulations is "if and only if". So what we have done is deduced that $$3x = 2x, \text{ for } x \neq 0 \text{ if and only if } 3=2.$$ Since the statement on the right is $\textit{always false}$, and since the connective is $\textit{if and only if}$ it means the statement on the left is $\textit{always false}$. Since $3x = 2x, x \neq 0$ is always false, we know that $3x \neq 2x$ for any $x$ other than 0. This means that the solution we found at the very beginning, $x = 0$, is the $\textit{only}$ solution, because that absurd conclusion of 3 = 2 that we got after dividing by $x$ means that no other real numbers can be solutions.

One also can probably easily see that in both of my examples, confusion can be avoided by factoring instead of dividing. E.g,

$$x^2 = 2x \implies x(x-2)= 0 \implies x = 0 \text{ or } x = 2,$$

$$2x = 3x \implies x(2-3) = 0 \implies x = 0.$$

But I solved my examples by dividing, since that was your specific question, to illustrate that it $\textit{is}$ ok to divide by the variable, as long as you know how to interpret the results of dividing by the variable.