general solution ,trig?

• Jun 1st 2009, 09:19 AM
oxrigby
general solution ,trig?
What is the general solution to: $\displaystyle 2sin2\Theta=\sqrt{3}tan2\Theta$? any ideas?
• Jun 1st 2009, 09:27 AM
Spec
$\displaystyle 2\sin 2\theta = \sqrt{3}\tan 2\theta = \sqrt{3}\frac{\sin 2\theta}{\cos 2\theta}\Longleftrightarrow \sin 2\theta = 0$ or $\displaystyle \cos 2\theta = \frac{\sqrt{3}}{2}$
• Jun 1st 2009, 09:33 AM
Amer
ok

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

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

dived the to side sin2 theta

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

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

$\displaystyle 2\theta=\frac{\pi}{6}....\theta=\frac{\pi}{12}+2n\ pi$
• Jun 1st 2009, 09:35 AM
oxrigby
would the end result be $\displaystyle \frac{\pi}{12}+n\pi$ or would it be $\displaystyle \frac{\pi}{12}+2n\pi$?
• Jun 1st 2009, 09:42 AM
Spec
You can't just divide away $\displaystyle \sin 2\theta$ from both sides since it's not a constant. If you do that, you'll miss out on all the solutions where $\displaystyle \sin 2\theta=0$

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

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

The general solution to $\displaystyle \cos x = C$ is $\displaystyle x=\pm v+2n\pi$ where $\displaystyle v$ is the angle and $\displaystyle n\in Z$
• Jun 1st 2009, 09:59 AM
dhiab
resolution
good night, I offer you a resolution
• Jun 1st 2009, 10:06 AM
dhiab
Quote:

Originally Posted by Amer
ok

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

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

dived the to side sin2 theta

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

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

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

good night, you not to pouver not to divide on sin 2theta
because sin2 theta can etre egal in zero
• Jun 1st 2009, 06:57 PM
Amer
I am sorry I should be careful when I answer questions (Doh)