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Okay, there's this simple equation I've been looking into for a while and I don't know why one way of solving it is not correct. See:

$$\sin(2x) + 3\cos(2x) = 0$$

Well, the most obvious would be to rearrange to get:

$$\tan(2x)=-3$$

and get the solutions from there, and this works. The problem comes in when another method is used:

$$\sin(2x) + 3\cos(2x) = 0$$ $$\sin(2x) + \dfrac{3\sin(2x)}{\tan(2x)} = 0$$ $$\sin(2x)\left(1 + \dfrac{3}{\tan(2x)}\right) = 0$$

Resulting in $$\sin(2x) = 0\ or\ \tan(2x)=-3$$

I cannot find anything wrong with this method, except that the solutions that $\sin(2x)$ give are not correct when I tested the results back in the original equation. There's probably something I'm overlooking, but I have to idea what. Could anyone explain? I tried to find an explanation to this on here and on the web but have failed so far.

[Of course, there is also the method of equating $\sin(2x) + 3\cos(2x) = \alpha\sin(x+\mu) = 0$, which confirms the first method, but I really want to know whether there's like an assumption or something I'm ignoring without knowing somewhere in the second method.]

EDIT: Thank you for the answers, but mentioning that dividing by zero is not possible it immediately made me think of something of the like of:

$$2x^2 - 5x = 0$$

If we take $x(2x - 5) = 0$, we get $x = 2.5$ or $x = 0$ which are both correct, but if we go with the 'first method' we get: $$2x^2 = 5x$$ $$2x = 5$$ $$x = 2.5$$ which ignores the existing solution $x = 0$.

reEDIT: Maybe what I'm trying to find is whether there is some way where I can 'see' that using this or that method will not be suitable, so I don't have to go back and reject excess solutions which are not valid?

Blue
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Jerry
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    The first step in "another method" is invalid when $\sin 2x=0$ (then you're dividing by $0$). – David Mitra Apr 07 '13 at 13:31
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    You similarly cannot go from $x+1=0$ to $x\cdot(1+\frac1x)=0$ and carelessly conclude that $x=0$ is a solution. – Hagen von Eitzen Apr 07 '13 at 13:40
  • @DavidMitra and HagenvonEitzen: I added an additional part to my section addressing another case: $2x^2-5x = 0$. I'm getting the feeling there is something with the trigonometry itself as opposed to the algebra handling (if I can call it like that?). – Jerry Apr 07 '13 at 13:51
  • I've removed the (division-by-zero) tag. Also, it is not entirely necessary to mark a topic as "Solved" in the title. Once you accept an answer (by ticking the check mark next to an answer), the votes box that appears next to the question whenever the question appears in a list (see, for example, to the right of this comment) will turn green, indicating that the question has received a satisfactory answer. – Willie Wong Apr 12 '13 at 08:21
  • Remember you can only divide both sides of $2x^2 = 5x$ by $x$ if $x \ne 0$. Hence you must first acknowledge that $x=0$ is a solution and then say, "Suppose $x \ne 0 \cdots$. – Steven Alexis Gregory Jul 03 '18 at 08:32

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From $$\sin(2x)\left(1+\frac{3}{\tan(2x)}\right)=0,$$ we conclude that $$\sin(2x)=0\quad OR\quad\tan(2x)=-3.$$ The former is not possible, since then $\tan(2x)=0$, and so $$1+\frac3{\tan(2x)}$$ isn't even defined. Hence, we have $\tan(2x)=-3,$ as you already found through other means.

Before you divide by an expression that may become $0$, you need to separate by cases: one case where that expression is $0$ (which may not be possible), and one where it is not $0$. Let me give you a few examples so you can get the idea.

$$x(x^2+1)=2(x^2+1)$$

Above, we split into the case that $x^2+1=0$ and the case that $x^2+1\neq 0$. In the latter case, division by $x^2+1$ shows us that $x=2$ without difficulty. In the former case, we must solve the equation $x^2+1=0$, which has no real solutions (complex solutions $\pm i$. Thus $x=2$ is the only real solution (and $\pm i$ are the other two complex solutions).

$$2x(x^2-1)=5(x^2-1)$$

Split into the case that $x^2-1=0$ and the case that $x^2-1\neq 0$. In the latter case, division by $x^2-1$ shows us that $x=\frac52$. In the former case, we need to solve $x^2-1=0$ (choose your favorite), and we'll find that $x=\pm1$. Hence, $x=\pm 1$ and $x=\frac52$ are the solutions.

Upshot: Before you divide, make sure that the expression you're dividing by isn't zero! If it can be zero, then deal with that separately.

Cameron Buie
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  • Does the same kind of 'procedure' (dividing by 0) not work in other circumstances like in $2x^2 - 5x = 0$? I updated my question with this following your answer. – Jerry Apr 07 '13 at 13:53
  • Indeed, in the case that $2x^2-5x=0,$ we see by factoring that $x=0$ or $x=\frac52.$ Removing the $x$ from the equation can only be achieved through division, but in order to do that, we must assume that $x\ne0$. From this, it would follow that $x=\frac52$, but we eliminated a solution in so doing. – Cameron Buie Apr 07 '13 at 13:56
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    I will add to my answer momentarily. – Cameron Buie Apr 07 '13 at 13:59
  • Hm, that's actually one way I've never seen before to go about solving an equation! Could you also perhaps include how you would split $2x^2-5x=0$ that way? I'm trying to rearrange it to fit the format $Ax(x^2\pm C)=B(x^2\pm C)$ but I'm not sure how to... – Jerry Apr 07 '13 at 14:17
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    We split $$2x^2=5x$$ into two cases: $x=0$ and $x\neq 0$. In the first case, there's no more work to be done. In the second, dividing by $x$ shows us readily that $x=\frac52$. The solutions are thus $x=0$ and $x=\frac52$. – Cameron Buie Apr 07 '13 at 14:23
  • Ohh haha. Thank you, I think that clears things up a bunch now :) – Jerry Apr 07 '13 at 14:27
  • One slight dissent: you aren't just dividing by zero, since this would need to be applied to every term on both sides of the equation. You're multiplying one term by 1. Unfortunately, the 1 becomes $\frac{0}{0}$ at times... – DJohnM Apr 07 '13 at 15:31
  • @User58220: Yes, and since those times involve division by zero, we must still split casewise. – Cameron Buie Apr 07 '13 at 15:57
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If $\sin(2x)=0$, then $2x= 0$ or $2x = \pi$. Therefore $\tan(2x) = 0$. And so you're not allowed to divide by $\tan(2x)$.

You can calculate the results with both ways, but you have to check the solutions afterwards.

Jakube
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  • I guess that's the simplest, but still quite tedious method xD. Thanks for your input +1 :) You might want to see the answer I picked which goes a little deeper than yours. – Jerry Apr 07 '13 at 14:29
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Hint:

$$\sin(2x)\left(1 + \dfrac{3}{\tan(2x)}\right) = 0$$

$\sin 2x =0 \implies \dfrac{\sin 2x}{\cos 2x}=0=\tan 2x$, which can't be a solution
of this expression: $$\sin(2x)\left(1 + \dfrac{3}{\tan(2x)}\right)=0$$

Inceptio
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  • Can't I use the argument 'Anything multiplied by 0 gives 0 so it doesn't matter what the other term is'? – Jerry Apr 07 '13 at 14:33
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In the step when writing $\cos(2x)$ as $\cos(2x) \frac{\sin(2x)}{\sin(2x)}=\frac{\sin(2x)}{\tan(2x)}$ you must asume that $\sin(2x) \neq 0$ so that you don't divide by zero. So in the last step the solution $\sin(2x)=0$ is not involved.

bergmeister
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  • Oops, I forgot to let you know that I updated my original post with another case where we actually do seem to be dividing by zero, which is $2x^2-5x=0$. – Jerry Apr 07 '13 at 14:34