Ask Uncle Colin: Some missing solutions

Dear Uncle Colin,

When I solve $2\tan(2x)-2\cot(x)=0$ (for $0 \le x \le 2\pi$) by keeping everything in terms of $\tan$, I get four solutions; if I use sines and cosines, I get six (which Desmos agrees with). What am I missing?

- Trigonometric Answers Not Generated - Expecting 'Nother Two

Hi, TANGENT, and thanks for your message!

Trig equations with $\tan$ in are often a bit subtle - just because of their tendency to go off to infinity at the drop of a hat.

Using $\tan$

My first thought would probably be to use the double-angle formula to get $\frac{4\tan(x)}{1 - \tan^2(x)} - \frac{2}{\tan(x)} = 0$ (*).

Multiplying through by $\tan(x)$ and $(1-\tan^2(x))$ gives $4\tan^2(x) - 2(1-\tan^2(x)) = 0$, or $6\tan^2(x) = 2$.

That leads to $\tan(x) = \pm \frac{\sqrt{3}}{3}$ and the four solutions $x = \piby6, \frac{5\pi}{6}, \frac{7\pi}{6}$ and $\frac{11\pi}{6}$.

Using $\cos$ and $\sin$

An alternative route is to work from (*) and deal with sines and cosines. Messily, we get $\frac{4 \frac{\sin(x)}{\cos(x)}}{1 - \frac{\sin^2(x)}{\cos^2(x)}} - \frac{2\cos(x)}{\sin(x)}$.

Tidy up the first fraction by multiplying top and bottom by $\cos^2(x)$:

$\frac{4 \sin(x) \cos(x)}{\cos^2(x)-\sin^2(x)} - \frac{2\cos(x)}{\sin(x)} = 0$

Converting that to a single fraction: $\frac{4 \sin^2(x)\cos(x) - 2\cos(x)(\cos^2(x)-\sin^2(x)}{\sin(x)(\cos^2(x)-\sin^2(x))}=0$

Now, the top of that is $4\sin^2(x)\cos(x) - 2 \cos(x)(\cos^2(x)-\sin^2(x))$, which we can obviously simplify. There's a factor of $2\cos(x)$, and the $\sin^2(x)$s can become $(1-\cos^2(x))$, which gives:

$\frac{2\cos(x) \left[ (2 - 2\cos^2(x)) - (2\cos^2(x)- 1)\right]}{\sin(x)(\cos^2(x)-\sin^2(x))} = 0$

$\frac{2\cos(x) \left[ 3 - 4\cos^2(x)\right]}{\sin(x)(\cos^2(x)-\sin(x))} = 0$

This is only true when either the bottom goes to infinity (which it never does) or if the top is 0, which occurs when $\cos(x)=0$ or $\cos(x) = \pm \frac{\sqrt{3}}{2}$. The second of those gives the same solutions as above, but the first gives additional solutions at $x = \piby 2$ and $\frac{3\pi}{2}$.

So what gives?

Did I foreshadow this? I think I might have done! I used a different method in the second part, where I didn't "multiply up" by something conceivably undefined.

Properly treated, (*) becomes $\frac{6\tan^2(x)-2}{\tan(x)\left(1-\tan^2(x)\right)} = 0$; as $x$ approaches $\piby2$ or $\frac{3\pi}{2}$, the bottom of the fraction is undefined - meaning those values also warrant further investigation.

Or does it?

We still need to check that the values we have are solutions to the original equation. If $x=\piby 2$, $2\tan(2x) - \cot(x) = 0 - 0$, which does indeed work. This sits poorly at first - after all, the working involves a $\tan(x)$ that's not defined at the relevant point! However, $\cot(x)$ most certainly is defined there. Writing (*) as $\frac{\cot(x)\left(6-2 \cot^2(x) \right)}{\cot^2(x)-1} = 0$ makes everything uncontroversially golden.

Hope that helps!

- Uncle Colin

* Edited 2017-07-19 to correct the denominator.
* Edited (twice) on 2017-07-19 after @realityminus3 spotted some lack of rigour in handling the 'infinite denominators' and an error in the algebra. Thanks, RM3!

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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