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My son asked for help with his maths homework last night. The question was to show that $$\tan (2x) = 5\sin(2x)$$ can be written as $$\sin(2x)(1-5\cos(2x))=0$$ My first response was to rearrange as $\tan (2x) - 5\sin(2x) = 0$, replace $\tan$ with $\frac{\sin}{\cos}$ and multiply through by $\cos$, etc. This worked fine. He then told me that he'd started by dividing by $\tan$ instead, and using the same replacement I'd used. This eventually gave him $1 - 5\cos(2x) =0$, ie, he'd lost part of the expression. My maths wasn't up to explaining why what he did was the wrong approach, so wondering if anyone can help me understand it? I kind of feel it might be something to do with dividing $\tan(2x)$ by $\tan(2x)$ to get 1 is throwing away information. Basically he'd like to understand why what he did was wrong, and what he should avoid doing going forward so as not to end up in a similar situation. He's asking his teacher today, but I'd like to understand it as well, hence the question here

Moo
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Kevin O'Donovan
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  • Dividing by $\tan$ is allowed if $\tan(2x) \ne 0$, or equivalently $\sin(2x) \ne 0$. – nicomezi Oct 16 '19 at 09:49
  • If you divide by a factor you lose some roots. If you divide $\sin (2x)(1-5\cos (2x))=0$ by $\sin (2x)$ you get $1-5\cos (2x)=0$ again. If you want to find all roots you should make sure not to divide by something which itself has some roots. – Kavi Rama Murthy Oct 16 '19 at 09:49
  • Thanks everybody, I feel I can understand this well enough to explain this to him now, and with hindsight it seems very clear. – Kevin O'Donovan Oct 16 '19 at 10:50

2 Answers2

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We can also divide by $\tan(2x)$ to obtain

$$\tan (2x) = 5\sin(2x) \implies 1-5\cos(2x)=0$$

but, in this case, we are implicitely assuming that $\tan(2x)\neq 0$.

The condition $\tan(2x)= 0$ needs finally to be checked a part in the original equation.

user
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You can only divide by $\tan(2x)$ if that term is not zero. Otherwise you divide by zero and that is not a valid transformation of the equation. The missing factor $\sin(2x)$ in his solution exactly covers the cases where $\tan(2x)=0$.

quarague
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