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 $$\displaystyle\arctan(-\sqrt3) = -\frac{\pi}{3}$$?
 A: 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.
A: Simply use that $\sin\frac{\pi}3=\frac{\sqrt3}2$ and $\cos\frac{\pi}3=\frac12$.
Then $\tan \frac{\pi}3=\dfrac{\frac{\sqrt3}2}{\frac12}=\sqrt3$.
Thus $\arctan \sqrt3=\frac{\pi}3$.
Adjust for the sign:  $\arctan -\sqrt3=-\frac{\pi}3$.
A: The fact that $\cos \frac{\pi}{3} = \frac{1}{2}$ is not something that requires memorization, but like many identities in mathematics, it is convenient and efficient to memorize because the proof is more sophisticated than the result.
In an equilateral triangle $\triangle ABC$, draw the altitude $\overline{AD}$ from $A$ to $\overline{BC}$.  Since $\angle A = \angle B = \angle C$ and their sum is $\pi$, it easily follows that $\angle B = \angle C = \pi/3$.  Since the altitude $\overline{AD}$ is by construction perpendicular to $\overline{BC}$, we conclude $\angle DAC = \angle DAB = \pi/6$, since $\angle ADC = \angle ADB = \pi/2$.  Consequently $\triangle ADC \cong \triangle ADB$ as they share $\overline{AD}$, thus  $BD = CD = AC/2 = AB/2$.  It follows that $$\cos \frac{\pi}{3} = \cos \angle C = \frac{CD}{CA} = \frac{1}{2}.$$
Then $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\tan \frac{\pi}{3} = \sqrt{3}$ follow from the Pythagorean theorem.

It is worth mentioning that the original equation $$\sin \pi x + \sqrt{3} \cos \pi x = 0$$ has another method of solution that uses the angle addition identity $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$  By dividing both sides by $2$, we obtain $$\frac{1}{2} \sin \pi x + \frac{\sqrt{3}}{2} \cos \pi x = 0.$$  Then recognizing the established identities above, we can write this as $$\cos \frac{\pi}{3} \sin \pi x + \sin \frac{\pi}{3} \cos \pi x = 0.$$  Applying the angle addition identity, we get $$\sin \left(\frac{\pi}{3} + \pi x\right) = 0,$$ from which it follows that $$\frac{\pi}{3} + \pi x = k \pi$$ for some integer $k$.  Therefore, $x = k - \frac{1}{3}$ for some integer $k$.  Since $x \in [0,2]$, the only such $k$ corresponding to $x$ in this interval are $k \in \{1, 2\}$.  This gives us $x \in \{2/3, 5/3\}$ as the complete solution set.
A: 
Given such triangle ABC, we have to prove $\angle ACB$ is $60$ degrees.

Construct $BD$ such that $BD=AD$.
Let $\angle BAC=\theta$.
Then, by isosceles triangles, $\angle DBA=\theta$.
Since $\angle ABC$ is a right angle, $\angle DBC=90-\theta$.
Applying sum of triangles on $\Delta ABC$, we get $\angle DCB=90-\theta$.
Hence $\Delta DBC$ is an isosceles triangle.

Let $AD=a$, $DC=b$.
By Pythagoras theorem, $AC=2$, hence $a+b=2$.
Considering isosceles triangle $DCB$, $DB=a$.
Considering isosceles triangle $DAB$, $DB=b$.
Thus $a=b=1$.
Therefore the three sides of $\Delta DBC$ are all equal to $1$ and it is an equilateral triangle. We get $\angle ACB=60$.

By definition, $\arctan(\sqrt 3)=\angle ACB$.
Noticing that $\arctan$ is an odd function and converting degrees to radians, 
$$\arctan (-\sqrt 3)=-\frac\pi 3$$
