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I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way: $$\tan(x) = 2\sin(x) \Longleftrightarrow \frac{\sin(x)}{\cos(x)} = 2\sin(x) \Longleftrightarrow \sin(x) = 2\sin(x)\cos(x) \Longleftrightarrow \sin(x) = \sin(2x)$$ But I'm getting a wrong result so I suppose that I can't do it in this way. Why?

This is my solution: $$\tan(x) = 2\sin(x) \Longleftrightarrow \frac{\sin(x)}{\cos(x)} = 2\sin(x) \Longleftrightarrow \sin(x) = 2\sin(x)\cos(x) \Longleftrightarrow \sin(x) = \sin(2x) \Longleftrightarrow \sin(2x) - \sin(x) = 0 \Longleftrightarrow 2\cos(\frac{3x}{2})\sin(\frac{x}{2}) = 0 \Longleftrightarrow \cos(\frac{3x}{2}) = 0 \vee \sin(\frac{x}{2}) = 0$$

Proper solution is $\cos(x) = \frac{1}{2} \vee \sin(x) = 0$

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I think he is trying to solve this equation; I am proceeding on that assumption. – ncmathsadist Mar 1 at 14:33
@paulgarrett - tdudzik is making use of the identity $$2\cos \theta\sin\phi = \sin(\theta+\phi) - \sin(\theta-\phi)$$ with $\theta = 3x/2$ and $\phi = x/2$ – Paul Sinclair Mar 1 at 17:38
@PaulSinclair, ok, and/but I more more largely baffled by the whole route taken, is the reason for my question. E.g., why not just take out the factor $\sin(x)$ in the first place, to have $\sin(x)\cdot (\cos x - 1)=0$ or whatever? All mysterious to me... – paul garrett Mar 1 at 17:51
@paulgarrett - certainly a much easier method, but this question was "what is wrong with this method?", not "how could I have done this more easily?". And there is actually nothing wrong with the path he chose, other than failing to go all the way. – Paul Sinclair Mar 1 at 18:09
@Shufflepants: If you do division by something that could be zero, that is illegal and punishable by contradiction. – user21820 Mar 2 at 5:19
up vote 12 down vote accepted

There is no problem. Your solution is correct, and the other solution is correct. But you need to do a little work to see that they give the same answers:

  • $\cos \frac {3x}2 = 0$ gives $x = \frac{(2k+1)\pi}3:\ x \in \{\pm \frac{\pi}3, \pm\pi,\pm \frac{5\pi}3, \pm \frac{7\pi}3, \pm 3\pi, \ldots\}$
  • $\sin \frac {x}2 = 0$ gives $x = 2k\pi:\ x \in \{0, \pm 2\pi, \pm 4\pi, \ldots\}$


  • $\cos x = \frac 12$ gives $x =\pm \frac{\pi}3 + 2k\pi:\ x \in \{\pm \frac{\pi}3, \pm \frac{5\pi}3, \pm \frac{7\pi}3, \ldots\}$
  • $\sin x = 0$ gives $x = k\pi:\ x \in \{0, \pm\pi, \pm 2\pi, \pm 3\pi,\pm 4\pi, \ldots\}$

Comparison of the values shows that the same ones are in both lists.

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You must disregard all points where $\cos(x) = 0$. These are outside of the domain of the initial expression you have on the left-hand side.

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Yeah, I know, but when I do this result will be correct? – tdudzik Mar 1 at 15:04
@tdudzik The result will be correct if there are any solutions where $\cos(x)\ne0$. – Simple Art Mar 2 at 0:07

The solutions to $\sin(\frac{x}{2})=0$ or $\cos(\frac{3x}{2})=0$ are given by $$\frac{x}{2} = \pi k \text{ or } \frac{3x}{2} = \frac{\pi}{2}+\pi k, k \in \mathbb Z,$$ which can be rewritten as $$x = 2\pi k \text{ or } x = \frac{\pi}{3}+\frac{2\pi}{3}k, k \in \mathbb Z.$$

The solutions to $\cos(x) = \frac{1}{2}$ or $\sin(x)=0$ are given by $$x = \frac{\pi}{3}+2\pi k, \frac{5\pi}{3}+2\pi k, \pi k, k \in \mathbb Z.$$ These are the same sets of points.

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Beat me by 5 seconds... – Paul Sinclair Mar 1 at 17:56
@PaulSinclair Hah, that does happen. +1 to you. :) – Dustan Levenstein Mar 1 at 18:01

Equations $\frac{\sin x}{\cos x} = 2\sin x$ and $\sin x =2\sin x\cos x$ are indeed equivalent since for $\cos x = 0$ you have $\sin x =\pm 1$, which doesn't give solution for the second equation.

So, to solve it, we have that either $\sin x = 0$ or $\cos x = \frac 1 2$, thus solutions are given by $x = 2k\pi$, $x = \pi + 2k\pi$, $x = \pm\frac\pi 3 + 2k\pi$, $k\in\mathbb Z$.

To summarize, there is no error in your manipulations.

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To solve, set $\sin(x) = \sin(2x) = 2\sin(x)\cos(x)$. Subtracting and factoring you get $$\sin(x)(2\cos(x) - 1) = 0.$$ This happens if $x = 0, \pi, \pi/3, 5\pi/3$. None of these is a solution to $\cos(x) = 0$, so they all work.

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I understand your solution but could you glance at my updated question and tell me what's wrong with my solution? – tdudzik Mar 1 at 17:01

There are two errors in your solution.

First, the transformation $$ \frac{\sin x}{\cos x} = 2\sin x \iff \sin x = 2\sin x \cos x $$ is false. The correct inference is $$ \left( \frac{\sin x}{\cos x} = 2\sin x \text{ and } \cos x \neq 0 \right) \text{ or } \cos x = 0 \iff (\sin x = 2\sin x \cos x \text{ and } \cos x \neq 0) \text{ or } \cos x = 0 $$ otherwise, going to the left, you divide by zero, and going to the right, you (might) delete the content of your equation by multiplying both sides by zero. (The hazards are $0 = 0 \implies 0 x = 0 y \not\Rightarrow x = y$ and $x \neq y \not\Rightarrow 0x \neq 0y \implies 0 \neq 0$.) Starting with either of your equations, you infer the entire other side of the more complicated bi-implication. Since this can introduce spurious solutions, all solutions must be checked in the original equation at the end of the process.

This suggests a general rule: Do not cancel. Instead subtract and factor. This highlights your second error. From $$ \sin x = 2 \sin x \cos x \text{,} $$ subtract and factor to get $$ (1-2\cos x)\sin x = 0 \text{.} $$ The first factor gives your $\cos x = \frac{1}{2}$. The second factor gives $\sin x = 0$. (This is because the product of several things being zero means at least one of them is.)

By proper inference, you should have \begin{align*} & & \tan x &= 2 \sin x \\ &\iff & \frac{\sin x}{\cos x} &= 2 \sin x \\ &\iff & \sin x &= 2 \sin x \cos x \text{ or } \cos x = 0 \\ &\iff & (1-2\cos x)\sin x &= 0 \text{ or } \cos x = 0 \\ &\iff & \cos x &= \frac{1}{2} \text{ or } \sin x = 0 \text{ or } \cos x = 0 \text{.} \end{align*} Checking for spurious solutions, all solutions of $\cos x = \frac{1}{2}$ and all solutions of $\sin x = 0$ are solutions of the given equation, but none of the solutions of $\cos x = 0$ are, so the final solution is $\cos x = \frac{1}{2} \text{ or } \sin x = 0$.

Somewhat shorter uses subtract and factor earlier, avoiding the complicated bi-implication altogether: \begin{align*} & & \tan x &= 2 \sin x \\ &\iff & \frac{\sin x}{\cos x} &= 2 \sin x \\ &\iff & \left( \frac{1}{\cos x} - 2 \right) \sin x &= 0 \\ &\iff & \frac{1}{\cos x} - 2 &= 0 \text{ or } \sin x = 0 \\ &\iff & \frac{1}{\cos x} &= 2 \text{ or } \sin x = 0 \\ &\iff & \cos x &= \frac{1}{2} \text{ or } \sin x = 0 \text{.} \end{align*}

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This strikes me as super confused. Your sequence of bi-implications after "By proper inference, you should have" are not correct at all; it should be "and $\cos x \neq 0$" after you've multiplied by $\cos x$. – Dustan Levenstein Mar 1 at 17:58
@DustanLevenstein : I could have "$(\sin x = 2 \sin x \cos x \text{ and } \cos x \neq 0) \text{ or } \cos x = 0$", exactly as in the prior bi-implication, but I invite you to show me how the set is changed by this complication. We cannot have either of $\cos x = 1/2 \text{ and } \cos x = 0$ or $\sin x = 0 \text{ and } \cos x = 0$. So the "$\dots \text{ and } \cos x \neq 0$" changes the set not at all. – Eric Towers Mar 2 at 23:57
What??? This is basic logic; "$\frac{\sin x}{\cos x} = 2 \sin x$" is equivalent to "$\sin x = 2 \sin x \cos x \text{ and } \cos x \neq 0$", not to "$\sin x = 2 \sin x \cos x \text{ or } \cos x = 0$" as you've said. I don't know why you nested the conditions on $\cos$; did you really misunderstand me that badly? – Dustan Levenstein Mar 3 at 1:01
Saying "... or $\cos x = 0$ introduces all solutions to $\cos x = 0$ to the set, none of which are solutions to the previous equation. – Dustan Levenstein Mar 3 at 1:03
@DustanLevenstein : "Since this can introduce spurious solutions, all solutions must be checked in the original equation at the end of the process." – Eric Towers Mar 3 at 1:07

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