# Solve $\sin(2\theta) -\tan(\theta) = 0 \$ for $0\leq \theta \leq 2\pi$

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 $\theta = \pi,2\pi$

• OK. Go ahead.${}$ – Andrés E. Caicedo May 10 '14 at 23:37
• @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 $\cos t =\frac{1-t^2}{1+t^2}$. – Jean-Claude Arbaut May 10 '14 at 23:52
• @MichaelT Your answer was good (with a minor sign mistake), and arguably more straightforward than the approach with tangents. Why delete it? – Jean-Claude Arbaut May 10 '14 at 23:58
• 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. – Jean-Claude Arbaut May 11 '14 at 0:06

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 $[0,2\pi]$.
You're missing a solution $$\sin(2x) = 2 \cos x \sin x = \tan x = \frac{\sin x}{\cos x} \implies \cos^2x = \frac{1}{2}$$ $$\therefore \cos x = \pm \frac{1}{\sqrt{2}} \quad or \quad \sin x = 0$$
• right! Forgot about the $\pm$ – Jeb May 11 '14 at 0:03