<s>[INST] 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 $\theta = \pi,2\pi$
 [/INST] You're missing a solution
$$ \sin(2x) = 2 \cos x \sin x = \tan x = \frac{\sin x}{\cos x} \implies  \cos^2x = \frac{1}{2} $$
$$\therefore \cos x = \pm \frac{1}{\sqrt{2}} \quad or \quad \sin x = 0 $$
</s>