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

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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}\sqrt{3}}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$2012-08-05

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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.

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    Thanks, That cleared it up. :)2012-08-05