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I want to use the fact that $$\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}$$

to solve $\sin(2\theta) -tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi$

My solution:

$\frac{2tan(\theta)}{1 + tan^2(\theta)} - tan(\theta) = 0 $

so $\frac{2tan(\theta) - (1 + tan^2(\theta)) tan(\theta)}{1 + \tan^2(\theta)} = 0$

so $ 2tan(\theta) - (1 + tan^2(\theta)) tan(\theta)= 0$

$ \implies \tan^3(\theta) + \tan(\theta) = 0$ $\implies \tan(\theta) [\tan^2(\theta) + 1] = 0$ $\implies \tan(\theta) = 0 \;\textrm{or}\; \tan^2(\theta) + 1 =0 $ since $\theta$ must be real.

Then we solve $\tan(\theta) = 0$ $\implies$ $\theta = n\pi, \ \ $ $ n \in Z \ \ $ so $\theta = \pi,2\pi$

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  • $\begingroup$ OK. Go ahead.${}$ $\endgroup$ – Andrés E. Caicedo May 10 '14 at 23:37
  • $\begingroup$ @MichaelT No, it's true, and it's quite usual: let $t=\tan \frac{\theta}{2}$, then $\sin \theta = \frac{2t}{1+t^2}$ and $\cos t =\frac{1-t^2}{1+t^2}$. $\endgroup$ – Jean-Claude Arbaut May 10 '14 at 23:52
  • $\begingroup$ @MichaelT Your answer was good (with a minor sign mistake), and arguably more straightforward than the approach with tangents. Why delete it? $\endgroup$ – Jean-Claude Arbaut May 10 '14 at 23:58
  • $\begingroup$ Your first so is of course wrong: you just introduced tangents to replace $\sin 2\theta$, but you write again the sine, instead of $2\tan \theta$. There is also a missing paren, but the line is wronge anyway. $\endgroup$ – Jean-Claude Arbaut May 11 '14 at 0:06
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Another way to solve:

Since identity $\sin (2\theta)=\dfrac{2\tan\theta}{1+\tan^2\theta}$, equation $\sin (2\theta)-\tan\theta=0$ is equivalent to \begin{align} \left(\dfrac{2}{1+\tan^2\theta}-1\right)\tan\theta&=0\;\;\;\text{ or}\\ \left(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\right)\tan\theta&=0\;\;\;\text{ or}\\ \frac{(1+\tan\theta)(1-\tan\theta)\tan\theta}{1+\tan^2\theta}&=0 \end{align} It follows $\tan\theta=\pm 1$ or $\tan\theta=0$. Thus $\theta=0,\frac{\pi}{4},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{7\pi}{4},2\pi$ are the solutions in $[0,2\pi]$.

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You're missing a solution $$ \sin(2x) = 2 \cos x \sin x = \tan x = \frac{\sin x}{\cos x} \implies \cos^2x = \frac{1}{2} $$ $$\therefore \cos x = \pm \frac{1}{\sqrt{2}} \quad or \quad \sin x = 0 $$

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  • $\begingroup$ right! Forgot about the $\pm$ $\endgroup$ – Jeb May 11 '14 at 0:03
  • $\begingroup$ Not a big mistake, and the answer is (I think) better than with tangents (faster, cleaner). ;-) $\endgroup$ – Jean-Claude Arbaut May 11 '14 at 0:03
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Try this \begin{align} \sin2\theta&=\tan\theta\\ \frac{2\tan\theta}{1 + \tan^2\theta}&=\tan\theta\\ \frac{2}{1 + \tan^2\theta}&=1\\ \tan^2\theta-1&=0\\ (\tan\theta-1)(\tan\theta+1)&=0. \end{align}

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tan squared theta = 1 can actually be turned into a quadratic equation and solved then replace back when you are done. Sorry can not edit answered this in a rush.

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