Math Help - general solution ,trig?

1. general solution ,trig?

What is the general solution to: $2sin2\Theta=\sqrt{3}tan2\Theta$? any ideas?

2. $2\sin 2\theta = \sqrt{3}\tan 2\theta = \sqrt{3}\frac{\sin 2\theta}{\cos 2\theta}\Longleftrightarrow \sin 2\theta = 0$ or $\cos 2\theta = \frac{\sqrt{3}}{2}$

3. ok

$2sin2\theta=\sqrt{3}tan2\theta$

$2sin2\theta=\sqrt{3}\left(\frac{sin2\theta}{cos2\t heta}\right)$

dived the to side sin2 theta

$2=\sqrt{3}\left(\frac{1}{cos2\theta}\right)$

$cos2\theta=\frac{\sqrt{3}}{2}$

$2\theta=\frac{\pi}{6}....\theta=\frac{\pi}{12}+2n\ pi$

4. would the end result be $\frac{\pi}{12}+n\pi$ or would it be $\frac{\pi}{12}+2n\pi$?

5. You can't just divide away $\sin 2\theta$ from both sides since it's not a constant. If you do that, you'll miss out on all the solutions where $\sin 2\theta=0$

It's similar to this equation: $x^3=x$ has obviously got one root $x=0$, but if you divide both sides with $x$ without taking that into consideration you'll end up with $x^2=1$, and thus miss a root.

The general solution to $\sin x = C$ is $x=v+2n\pi$ or $x=\pi-v+2n\pi$ where $v$ is the angle and $n\in Z$

The general solution to $\cos x = C$ is $x=\pm v+2n\pi$ where $v$ is the angle and $n\in Z$

6. resolution

good night, I offer you a resolution

7. Originally Posted by Amer
ok

$2sin2\theta=\sqrt{3}tan2\theta$

$2sin2\theta=\sqrt{3}\left(\frac{sin2\theta}{cos2\t heta}\right)$

dived the to side sin2 theta

$2=\sqrt{3}\left(\frac{1}{cos2\theta}\right)$

$cos2\theta=\frac{\sqrt{3}}{2}$

$2\theta=\frac{\pi}{6}....\theta=\frac{\pi}{12}+2n\ pi$
good night, you not to pouver not to divide on sin 2theta
because sin2 theta can etre egal in zero

8. I am sorry I should be careful when I answer questions