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 $\theta = \pi,2\pi$

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 $[0,2\pi]$.
