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

Question

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 $\theta = \pi,2\pi$

Answer 1

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 $[0,2\pi]$.

Answer 2

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

Answer 3

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}

Answer 4

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.