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I have this equation

$$\arctan(x)+\arctan(2x)=\frac{\pi}{3}$$

that ends up with two roots but when I graph the equation online the graph only intercepts the x-axis once. So where is my problem?

$$\begin{align} \arctan(x)+\arctan(2x) &=\frac{\pi}{3} \tag{1}\\ \tan(\arctan(x)+\arctan(2x)) &=\tan\left(\frac{\pi}{3}\right) \tag{2} \\ \tan(\arctan(x)+\arctan(2x)) &=\sqrt3 \tag{3} \\ \frac{\tan(\arctan(x))+\tan(\arctan(2x))}{1-\tan(\arctan(x)\cdot\arctan(2x))} &=\sqrt3 \tag{4} \\ \frac{3x}{1-2x^2}&=\sqrt3 \tag{5} \\ 3x &=\sqrt3\cdot(1-2x^2) \tag{6} \\ 3x &=\sqrt3-2\sqrt3x^2 \tag{7} \\ 2\sqrt3x^2+3x-\sqrt3 &=0 \tag{8} \end{align}$$

Now I look for the roots using $$b^2-4ac=9-(-4\cdot2\sqrt3\cdot\sqrt3)=33 \tag{9}$$ $$x_1= \frac{-3+\sqrt{33}}{4\sqrt3} \tag{10}$$ $$x_2= \frac{-3-\sqrt{33}}{4\sqrt3} \tag{11}$$

I hope this is right (not sure).

But If it is, I don't understand why graphing this function online gives me only one root, which corresponds to $x_1$.

enter image description here

I also used a website allowing to solve trigonometric equations on the fly in order to check my answer and same result there: their answer is unique and corresponds to my $x_1$ only. Can someone explain this to me please?

Thanks for your help.

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2 Answers 2

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This happened because not all steps in your solution are equivalences. One particular step (I'll tell you which one in a moment) is an implication, but not an equivalence — and it's there where we may acquire extraneous solutions; and apparently in this case we did.

To start with a more basic example: think about squaring both sides of an equation. If $a=b$, then $a^2=b^2$. But the converse is not necessarily true: $a^2=b^2$ does not imply $a=b$. For example, $(-5)^2=5^2$, and yet $-5\neq5$. This leads to common mistakes when students solve equations and square along the way: $$x=-\sqrt{x} \implies x^2=\left(-\sqrt{x}\right)^2 \implies x^2=x \implies x=0,1,$$ and yet only $x=0$, but not $x=1$, is a solution to the original equation.

The same thing happened here: if $a=b$, then $\tan(a)=\tan(b)$; but the converse is not necessarily true: $\tan(a)=\tan(b)$ does not imply $a=b$. For example, $\tan(0)=\tan(\pi)$, and yet $0\neq\pi$.

You effectively solved the equation $\tan(\mathrm{LHS})=\tan(\mathrm{RHS})$, and it has two solutions. But there's no guarantee that any of them satisfies the original equation. You need to check each of them by plugging into the original equation.

An alternative is to determine some constraints on the values of $x$ to guarantee that this step is reversible. Sometimes it is a better idea. But I'm not sure if it's easy enough here; so maybe plugging in is what you need to do.

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  • $\begingroup$ nice answer.... $\endgroup$ May 20, 2020 at 16:24
  • $\begingroup$ Thank you. I understand your logic. it makes total sens. I just don't see where, in particular, in my calculation, I get rid of both tan's. I elaborate from that but I never get rid of them. So, to me, technically, I added them (which you say is ok) but don't get them out. But ok, the principle is understood. Now, I don't see where and how I should correct this in order to be accurate, without writing a text under the calculation. Maths should be enough but I don't know how to get there. I understood alright but don't see how to correct this. $\endgroup$ May 20, 2020 at 17:03
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    $\begingroup$ @BachirMessaouri: It was the step from (1) to (2), where you take $\tan$ of both sides of the equation, which potentially results in introducing extra roots. In other words, equations (1) and (2) are not equivalent: (1) implies (2), but not vice versa. Effectively you solved equation (2), which is $\tan(\mathrm{LHS})=\tan(\mathrm{RHS})$, and you obtained two roots $x_1,x_2$, which are the roots of equation (2). But then you say that the same numbers are roots of equation (1), $\mathrm{LHS}=\mathrm{RHS}$, which is not necessarily true. $\endgroup$
    – zipirovich
    May 20, 2020 at 18:43
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    $\begingroup$ @BachirMessaouri: In other words, your getting rid of tans was logical, not computational. It happened when from the correct statement "$x_1,x_2$ are the roots of equation (2)" you made an incorrect conclusion that "$x_1,x_2$ are the roots of equation (1)". $\endgroup$
    – zipirovich
    May 20, 2020 at 18:44
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    $\begingroup$ @BachirMessaouri: Note that I didn't say that you did anything wrong in your process. Your only mistake was not completing your process; you jumped to a logically incorrect conclusion instead. Performing such one-way implication steps is normal and often unavoidable. The key to doing this correctly is to clearly realize when you perform such one-way steps, and then to remember to verify the roots in the end to filter out the extra ones. So you don't need to change anything in your solution, but you must add verification by plugging in. $\endgroup$
    – zipirovich
    May 20, 2020 at 18:47
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Take tan on both sides of equation. The trig problem has been designed to turn quadratic:

$$\dfrac{3x}{1-2x^2}= {\sqrt 3}; \quad 2 x^2+\sqrt 3 x -1=0 $$

It is a quadratic equation whose discriminant $\Delta^2= 3-4\cdot 2\cdot(-1) =11>0, \,$ so must have two roots.

All your work is correct. But plotting and calculating inverse functions entails loss of a part of solution almost always.

The arctan function accepts $ \pm \pi $. To capture all roots we should consider concurrency of the three curves ($ \tan^{-1}$, parabola) with the x-axis. The second arctan was missing in your graph.

Making a more comprehensive plot with coterminal $\pi$ addition we have the full field access to include all intersections.

ArcTan Co Terminal Soln

The roots are $(x_1,x_2)\approx (-1.26,+0.396)$ considering coterminal angles, in full agreement with the two roots of parabola.

Good problem.

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  • $\begingroup$ Thank you very much! I wasn't even aware of the implications of those roots and the "three curves". It's very interesting. I will dig deeper into this as I clearly missed a point here. $\endgroup$ May 20, 2020 at 19:48
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    $\begingroup$ Welcome. While seeking solutions in Mathematica CAS a message is routinely issued... Inverse functions occur, so something might be missing etc. This is one such representative case. $\endgroup$
    – Narasimham
    May 20, 2020 at 20:30

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