What is the general solution to: $\displaystyle 2sin2\Theta=\sqrt{3}tan2\Theta$? any ideas?
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 $
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 $\displaystyle n\in Z$
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2shared - download R1.pdf