Trig or treat

the main problem with this problem is that it looks tremendously horrible at first sight. i cannot find any good method to solve this. if you have, please do comment

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First we write $3x=Z$ .. (for simplicity, obviously)

then we write $sec^{2}Z=1+tan^{2}Z$ , bring all the terms to the L.H.S, group them, to get our equation:

$\sqrt{3}tan^{3}Z+(2-\sqrt{3})tan^{2}Z-(2+\sqrt{3})tanZ+\sqrt{3}=0$

From here, it is very easy to observe that $tanZ=1$ satisfies the equation.

Thus, by good-ol' factorisation again and again, we get :

$(tanZ-1)(tanZ+\sqrt{3})(tanZ-\frac{1}{\sqrt{3}})=0$

From there, calculation yields the 3 lowest positive values of $x$ as:

$x=10^{\circ},15^{\circ},40^{\circ}$

Our sum $=\boxed{65}$