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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
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    $\begingroup$ The problem is that your expression is undefined if $\cos(\theta)$, $\cos(2\theta)$ or $\cos(3\theta)$ is $0$. $\endgroup$ – Robert Israel Feb 21 '17 at 7:07
  • $\begingroup$ The same question has been asked by @Siddhant. Check this out math.stackexchange.com/questions/1878004/… $\endgroup$ – Saksham Feb 21 '17 at 17:21
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You could have solved the problem using the tangent multiple angle formulae.

Using $t=\tan(\theta)$ $$\tan(2\theta)=\frac{2t}{1-t^2}\qquad \tan(3\theta)=\frac{3t-t^3}{1-3t^2}$$ This makes, after some minor simplifications

$$\tan \theta + \tan 2\theta + \tan 3\theta - \tan \theta \tan 2\theta \tan 3\theta=\frac{2 t \left(3-t^2\right)}{1-t^2}=0$$ and then the solutions assuming that ${1-3t^2}\neq 0$ and ${1-t^2}\neq 0$.

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your intuition should be correct here's why

$\left(\frac{\sin\theta}{\cos\theta}+\frac{\sin2\theta}{\cos2\theta }+\frac{\sin3\theta}{\cos3\theta}=\frac{\sin\theta\sin2\theta\sin3\theta}{\cos\theta\cos2\theta\cos3\theta}\right)*\cos\theta\cos2\theta\cos3\theta$

$\Rightarrow \sin\theta\cos2\theta\cos3\theta+\cos\theta\sin2\theta\cos3\theta+\cos\theta\cos2\theta\sin3\theta= \sin\theta\sin2\theta\sin3\theta$

$\Rightarrow \cos\theta(\sin2\theta\cos3\theta+\cos2\theta\sin3\theta )+\sin\theta(\cos2\theta\cos3\theta-\sin2\theta\sin3\theta)=0 $

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

$\Rightarrow sin([2\theta+3\theta]+\theta)=0 $

$\Rightarrow sin(6\theta)=0 $

So, I'm uncertain that your answer is wrong...

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We have $$\tan (A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1-(\tan A \tan B + \tan B \tan C + \tan C \tan A)}$$ and the given condition implies that $\tan (6\theta) = 0$. Thus $\theta = \frac{n\pi}{6}$, $n \in \mathbb{Z}$

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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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