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?


  • 4
    $\begingroup$ 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. $\endgroup$
    – Valtteri
    Feb 14, 2013 at 21:58
  • 2
    $\begingroup$ 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 $\sin 5z=-1$. $\endgroup$ Feb 14, 2013 at 21:59
  • $\begingroup$ @BrianM.Scott Yes, you're right, thanks. $\endgroup$
    – Julien
    Feb 14, 2013 at 22:03

1 Answer 1


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


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