Solve the equation 3 sin(x)^2 = cos(x)^2, for 0° ≤ x ≤ 180°

Could someone please explain how to do this? I have the answer (I'm reviewing for an upcoming exam) but it doesn't help me much.

Answer:
tan2x = 1/3
--> tanx = ± 1/√3
--> x = 30° or x = 150°

To avoid later confusion, I'm changing the name of the argument to θ
3 sin²θ = cos²θ

Divide both sides by the RHS:
3(sinθ/cosθ)² = 1

But sinθ/cosθ = tanθ, so
3 tan²θ = 1
tan²θ = ⅓

So
tanθ = ±√⅓ = ±1/√3

Now draw a circle centered at the origin in the xy-plane, from an angle of 0º to 180º. Tangent is y/x.
So take y = 1, x = √3 to be on that circle, in the 1st quadrant. Then the radius, r, to that point is given by
r² = x² + y² = 3 + 1 = 4
r = 2
So y/r = ½ ::: but y/r = sinθ, and you should recognize

θ = 30º from that.

Then for the minus value,
take y = 1, x = -√3 [the minus has to go onto the x in order to stay in that upper ½-circle],
and you can see that this just puts the point at the mirror reflection, to the left of the y-axis, of the first point, and that makes the angle

θ = 180º - 30º = 150º

which, BTW, is the other angle whose sine is ½.

Note that tan(x) = sin(x)/cos(x). So by dividing both sides over cos^2 (x),

3* [sin(x)/cos(x)]^2 = 1 ----> 3* tan^2 (x) = 1 ----> tan^2 (x) = 1/3 --->tan(x) = ± 1/√3
----> x = 30° or x = 150°.

Hope this help.
Best Regards.