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I am given this equation to solve for $\theta$ :

$$\tan2\theta = 3\cot\theta$$

And I am given the following answer:

$$\tan\theta = \pm\sqrt{\frac{3}{5}}\\ \text{or denominator} = \infty\\ \theta = 37.8°, 218°, 142°, 322°, 90°, 270°$$

What does it mean by denominator = $\infty$, and where did $90°$ or $270°$ come from?

Edit:
This is the full working out
enter image description here

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  • $\begingroup$ See now, please. I posted a right reasoning. $\endgroup$ Jun 6 '17 at 17:02
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The domain of given equation is $\{\theta|\cos{2\theta}\sin{\theta}\neq0\}$.

We need to solve $$\frac{\sin2\theta}{\cos2\theta}=\frac{3\cos\theta}{\sin\theta}$$ or $$2\sin^2\theta\cos\theta-3\cos\theta(2\cos^2\theta-1)=0$$ or $$\cos\theta(2-2\cos^2\theta-6\cos^2\theta+3)=0,$$ which gives $\cos\theta=0$, which is $\theta=90^{\circ}+180^{\circ}k$ for $k\in\mathbb Z$ or $$\cos^2\theta=\frac{5}{8},$$ which is $$\frac{1}{\cos^2\theta}=\frac{8}{5}$$ or $$1+\tan^2\theta=\frac{8}{5}$$ or your $$\tan\theta=\pm\sqrt{\frac{3}{5}}$$

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  • $\begingroup$ I didn't think of it this way. But this makes things clearer as well, thanks. Could you explain what $k\in\mathbb Z$ means in this case? I don't understand that notation. $\endgroup$
    – Kyzen
    Jun 6 '17 at 17:05
  • $\begingroup$ @Kyzen It just says that $k$ is an integer number. $\endgroup$ Jun 6 '17 at 17:08
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When $\theta = 90°, 270°$ then $\tan\theta = \infty$, this implies that the denominator is $\infty$ on line 4 of the calculations. Which also implies that the equation itself is satisfied (because the numerator is of order $\tan^3\theta$ and numerator only of order $\tan^2\theta$).

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  • $\begingroup$ Oh, thanks! The remaining question is how from $tan\theta = \pm\sqrt{\frac{3}{5}}$ is $\text{denominator} = \infty$ obtained? $\endgroup$
    – Kyzen
    Jun 6 '17 at 16:21
  • $\begingroup$ These two solutions are independent. Or you make the denominator $\infty$ or you make the numerator $0$. $\endgroup$
    – NDewolf
    Jun 6 '17 at 16:22
  • $\begingroup$ Thanks, but I don't still understand where $\text{denominator} = \infty$ comes from. Is there something special about the question asked that allows for the denominator to be infinite or the numerator to be 0? $\endgroup$
    – Kyzen
    Jun 6 '17 at 16:25
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    $\begingroup$ No, thats just the solution to the equation. There are two possibilities: or the numerator on line 4 is 0, or the denominator is $\infty$. There are no other ways to make the fraction equal to 0. $\endgroup$
    – NDewolf
    Jun 6 '17 at 16:27
  • $\begingroup$ Ah, I finally get it. I'm guessing that by multiplying the denominator on line 4 to get rid of it and to let only the numerator remain, I am losing a solution? $\endgroup$
    – Kyzen
    Jun 6 '17 at 16:31

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