[Math] Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $

trigonometry

I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in Z $)

Here is what I've done

$$\frac {\sin \theta}{\cos \theta} + \frac {\sin 2\theta} {\cos 2\theta} + \frac{\sin 3\theta}{\cos 3\theta}= \frac {\sin \theta}{\cos \theta} \frac {\sin 2\theta} {\cos 2\theta} \frac{\sin 3\theta}{\cos 3\theta}$$

$$ \frac {\sin \theta \cos 2\theta \cos 3\theta + \cos \theta \sin 2\theta \cos 3\theta + \cos \theta \cos 2\theta \sin 3\theta – \sin \theta \sin 2\theta \sin 3\theta }{\cos\theta \cos 2\theta \cos 3\theta} = 0 $$

$$\cos 2\theta \{\sin\theta \cos 3\theta + cos \theta \sin 3\theta \} + \sin 2\theta \{\cos \theta \cos 3\theta – \sin \theta \sin 3\theta \} = 0 $$

$$\cos 2\theta \sin(3\theta + \theta) +\sin2\theta \cos(3\theta + \theta) = 0 $$

$$ \cos 2\theta \sin 4\theta + sin 2\theta cos 4\theta = 0$$

$$ \sin (2\theta + 4\theta) = 0$$

$$\sin 6\theta = 0 $$

$$ \theta = \frac {n\pi}{6}, n \in Z$$

I understand from this question that whatever mistake I am making is in the third step, where I remove $\cos \theta \cos 2\theta \cos 3\theta $ from the denominator. However, despite reading through the aforementioned post, I couldn't really get the intuition behind why this is wrong.

I'd like :

  1. To understand the intuition behind why removing $\cos \theta \cos 2\theta cos 3\theta $ is a mistake.
  2. To know how to solve this question correctly
  3. How do I avoid making these types of mistakes when solving trigonometric equations

Best Answer

In your simplified expression there is $\cos 3 \theta $ term in the denominator that goes to zero for solution you obtained. Should be checked before accepting or discarding it as a valid solution.

I cannot resist an elementary trig approach ..

If $ (A+B+C) = 2 \pi, $ then $ {(\tan A + \tan B + \tan C) = \tan A \tan B \tan C} $

If $ {(\tan A + \tan B + \tan C) = \tan A \tan B \tan C} $ then $(A+B+C) = 2 \pi $

is among possible solutions.

In the above if $ A= t, B=2t, C=3t$ then $t= \pm 2 \pi/6= \pm \pi/3,\pm 2\pi/3, ... $

By inspection $ t= 0 ,\,2 k\pi,$ plus co-terminals

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