I'm going through Khan Academy and I'm stuck at Trig Identity and there is something I don't understand.

Given that $x$ is in the first quadrant and $\sec x$ is $\frac{2\sqrt3}3$, what is $\cos x$?

$\sec x=\frac1{\cos x}$

$\sec x=\frac{2\sqrt3}3$

$\frac1{\cos x}=\frac{2\sqrt3}3$

$\cos x=\frac{\sqrt 3}2$

As we can see, this is how they say it should be done, but I don't understand how they went from step 3 to step 4. Can someone please explain to me.

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$\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}$ – lab bhattacharjee Aug 5 '12 at 9:24
@labbhattacharjee Do you think you can break it down into a few steps. I don't consider my self to be an expert at math (which is why I'm going through Khan Academy) so I would appreciate it if you can break it down into steps for me. – Caesar Aug 5 '12 at 9:28
$\frac{3}{2 \sqrt3}=\frac{\sqrt3 \times \sqrt3}{2 \sqrt3}=\frac{\sqrt3}{2}$ – Saurabh Aug 5 '12 at 9:56
$\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}\sqrt{3}}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$ – Ravi Aug 5 '12 at 9:57

Start with $${1\over\cos x}={2\sqrt3\over3}$$ Multiply both sides by $\cos x$: $$1={2\sqrt3\over3}\cos x$$ Multiply both sides by 3: $$3=2\sqrt3\cos x$$ Divide both sides by $2\sqrt3$: $${3\over2\sqrt3}=\cos x$$ Now proceed as in the comments.