HoW dO I dO tHiS pRoBleM?

Question

So I am in Honors PreCal and we have a test. I'm reviewing the problems. I have no idea how to do this:
(tan^2 3x) = 3
my teacher said to square root both sides but I have no idea where to go from there!?!?!?!? Help!

Answer 1

(tan^2 3x) = 3
So tan 3x = sqrt (3)
3x = arctan sqrt(3)
x = (arctan sqrt(3))/3
x=0.34906585

Answer 2

Let u = 3x. Then x = u/3.

Now the problem simply becomes tan² u = 3, and tan u = ±√3. We can use both the positive and negative value for the square root of 3 since both when squared yield 3.

Now all we have to do is find the angles whose tangents equal +√3 and -√3, then divide those angles by 3.

The angle whose tangent equals √3 happens to be 60°. So u = 60° and x = u/3 = 20°.

The angle whose tangent is equal to -√3 is 300°. So, u in this case is 300°, and x = u/3 = 100°.

So, your answers are 3x = 60° ----> x = 20° and 3x = 300° ----> x = 100°.

Answer 3

I don't know. But WhY dO yOu typE LIke THIs?