[Math] KhanAcadethe-Trig Identities 1

algebra-precalculustrigonometry

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

enter image description here

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.

Best Answer

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.

Related Question