general solution ,trig?

**oxrigby** · June 1st 2009, 09:19 AM

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

**Spec** · June 1st 2009, 09:27 AM

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

**Amer** · June 1st 2009, 09:33 AM

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

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

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

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

**oxrigby** · June 1st 2009, 09:35 AM

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

**Spec** · June 1st 2009, 09:42 AM

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 $n\in Z$

**dhiab** · June 1st 2009, 09:59 AM

good night, I offer you a resolution
2shared - download R1.pdf

**dhiab** · June 1st 2009, 10:06 AM

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

**Amer** · June 1st 2009, 06:57 PM

I am sorry I should be careful when I answer questions

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