Proving $\arctan (-\sqrt3) = \frac{2\pi}{3}$ using basic methods Background
A recent exam I'm looking at, has the following trigonometric equation:

$$\sin(\pi x) + \sqrt3 \cos(\pi x) = 0, \quad x \in [0, 2]$$

After a simple re-write, we get $$\tan(\pi x) = -\sqrt 3$$
Note, on this exam, the student doesn't have any tools or sheets available. They may or may not remember the inverse tangent here, and let's assume they do not.
Question
What are some solid ways that a student at this level, knowing basic trigonometry, could reason their way to the right answer here?
My thoughts
In a $30-60-90$ triangle, we soon discover that $\tan(60^\circ) = \sqrt3$, but this requires memorizing the ratios of the sides in a $30-60-90$. From here, we can use the unit circle to look at where the tangent function is positive and negative, and we would find the answer.
Not entirely unreasonable, but if we assume they remember this, they may remember the inverse tangent just as well.
Are there any better, more reasonable ways a student could be expected to find that $$\displaystyle\arctan(-\sqrt3) = -\frac{\pi}{3}$$?

I agree with one of the comments to your question that it’s worth memorizing some of the basic triangles and the associated sines and cosines. However, a 30-60-90 triangle can be quickly reverse-engineered from $\tan\theta = -\sqrt3$: we have $$\frac yx = -\sqrt3 \\ x^2+y^2=1$$ from which $4x^2=1$, so $x=\pm\frac12$ and $y=\mp\frac12\sqrt3$. The hypotenuse is, of course, $1$. I would hope that a student at that level could recognize either of the resulting triangles as half an equilateral triangle.
