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Solve: $\arctan(2x)+\arctan(3x) = \frac{\pi}{4}$

I started by applying tan on both sides,

$\frac{2x+3x}{1-2x\times3x}=\tan\frac{\pi}{4}$

This yields $x=\frac16,$ $x=-1$

But -1 doesn't satisfy the equation! Can you help me identify my mistake? Thanks.

P.S. I saw some similar questions posted on this same problem. But it didn't address how to correctly solve this equation.

Michael Rozenberg
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emil
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3 Answers3

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Your argument only tells you that if $x$ is a solution of the equation then $x$ must be one of the two values you have obtained. It doesn't not follow that both the values actually satisfy the original equation. (If you try to retrace your steps you will have to use that fact $\tan$ is a one-to-one function, but this is not true). After finding possible values of $x$ you should go back to the original equation and keep only those values that actually satisfy it.

Kavi Rama Murthy
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Adding angles corresponds to multiplying complex numbers.

$\arctan(\theta)$ is the angle (argument) of the complex number $1+\theta i.$ Then

$(1+2xi)(1+3xi) = (1-6x^2)+ 5xi$

is the complex number with angle

$\arctan(2x)+\arctan(3x)$.

Therefore we want to find $x$ such that $(1-6x^2)+5xi$ is a complex number with angle $\pi / 4$. In other words we want it to be a real multiple of $1+i$, or equivalently a complex number with equal real and imaginary part. However, not all such numbers have angle $\pi/4$ so we can obtain false solutions.

This leads to the same equation that you have, although now it may be easier to see that for $x < 0$, you are adding angles in the clockwise direction, i.e. your angles are negative. Therefore $x = -1$ gives a false solution, and corresponds to the complex number $-5-5i$ that has angle

$\arctan(-2)+\arctan(-3) = -3\pi/4$.

Gunnar Sveinsson
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$-1$ is not valid because $\arctan(-2)+\arctan(-3)=-135^{\circ}\neq45^{\circ}.$

We got this value because $\tan45^{\circ}=\tan(-135^{\circ})$

Michael Rozenberg
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