# Is it allowed to divide an equation by an expression which can be equal to zero?

I need a help in such a problem and will greatly appreciate any suggestions.

I was taught, division of an equation by an expression which can be equal to zero can lead to missing roots. But I thought that this is not forbidden completely, cause always in such a situation I simply solved additional equation (which was a divider) and checked if its roots are roots of an initial equation. For example:

$$(x + 2)^2 + x + 2 = 0$$

Dividing this by $( x + 2 )$ gives

$$x + 2 + 1 = 0$$

so the root is $x = -3$, then I simply check whether the $x + 2 = 0$ gives a root which fits the initial equation, and it does: $x = -2$.

I do not have a great experience in mathematics, but I solved a couple of hundreds of equations more complex than this and never had a trouble. But today I encountered this issue. Here is the equation:

$$3\sin(5z) -2\cos(5z) = 3$$

I saw that $\cos(5z) = 0$ is the one of the solutions, but I divided all by it and got:

\begin{align} 3\tan(5z) - 2 &= 3 \\ 3\tan(5z) &= 5 \\ \tan(5z) &= \frac{5}{3} \\ 5z &= \arctan(\frac{5}{3}) + \pi k , k \in \mathbb{Z} \\ z &= \frac{\arctan(\frac{5}{3}) + \pi k}{5}, k \in \mathbb{Z} \end{align}

the second root is:

\begin{align} \cos(5z) &= 0 \\ z = \frac{\frac{\pi}{2} + \pi k}{5} &= \frac{\pi (1 + 2 k)}{10}, k \in \mathbb{Z} \end{align}

But the first one is not a real root for some reason. This was the first thing which confused me here, I knew that I can lose a root but I never found parasite ones in similar situations. I was taught that parasite roots can appear only when doing a multiplication by an expression which can be equal to zero or an exponentiation to an even exponent, not in the cause of a division.

The second confusion is the fact that I actually have lost some roots, cause the right solution from the textbook was (with the introduction of the so called auxiliary angle $\sqrt{3^2 + 2^2} = \sqrt{13} \implies \gamma = \arcsin(\frac{3}{\sqrt{13}})$):

\begin{align} \underbrace{\frac{3}{\sqrt{13}}}_{\sin(\gamma)} \ sin(5z) - \underbrace{\frac{2}{\sqrt{13}}}_{\cos(\gamma)} \ cos(5z) &= \frac{3}{\sqrt{13}} \\ \sin(\gamma)\sin(5z) - \cos(\gamma)\cos(5z) &= \frac{3}{\sqrt{13}} \\ \cos(\gamma + 5z) &= -\frac{3}{\sqrt{13}} \\ z &= \pm \frac{1}{5}\arccos(-\frac{3}{\sqrt{13}}) - \frac{1}{5}\arcsin(\frac{3}{\sqrt{13}}) + \frac{2\pi k}{5} \end{align}

So I actually have two questions:

1. Is it possible to lose roots permanently when dividing an equation by an expression which may be equal to zero ?
2. Is it possible to gain parasite roots in such a case? Or in which situations they can appear in general, not only concerning division?

Thanks.

• Instead of dividing, you should factorize. IE: the first equation is $(x+2)(x+2+1)=0$. That makes the "division by zero" problem disappear. – Valtteri Feb 14 '13 at 21:58
• There’s another problem besides the algebra error that Ludolila pointed out: $\cos 5z=0$ is not a solution. $5z=\left(2n+\frac12\right)\pi$ is a set of solutions making $\cos 5z=0$ and $\sin 5z=1$, but $5z=\left(2n+\frac32\right)\pi$, which also makes $\cos 5z=0$, is not a set of solutions, because it makes $\sin 5z=-1$. – Brian M. Scott Feb 14 '13 at 21:59
• @BrianM.Scott Yes, you're right, thanks. – Julien Feb 14 '13 at 22:03

You should have divided both sides of the equation by $\cos (5z)$, and you only divided the left side.