Solve the following equation:

$$
\sin x \tan x=\frac{1}{2 \sqrt{3}}
$$
Our equation can also be written as:

$$
\frac{\sin ^{2} x}{\cos x}=\frac{1}{2 \sqrt{3}}
$$

or

$$
\frac{1-\cos ^{2} x}{\cos x}=\frac{1}{2 \sqrt{3}}
$$

after rearranging:

$$
\cos ^{2} x+\frac{1}{2 \sqrt{3}} \cos x=1
$$

from which

$$
\cos x=\frac{-1 \pm 7}{4 \sqrt{3}}
$$

If we consider the lower sign, then the absolute value of the fraction is greater than $1 \mathrm{~s}$, so only the upper sign provides a valid solution; hence

$$
\cos x=\frac{3}{2 \sqrt{3}}=\frac{1}{2} \sqrt{3}
$$

thus

$$
x=30^{\circ}
$$

or if $k$ denotes any positive integer, then

$$
x=k \cdot 360^{\circ} \pm 30^{\circ}
$$

(Károly Oltay.)

The problem was also solved by: Andráschek F., Barabás S., Bobál S., Breuer M., Csete A., Dolowschiak M., Filkorn J., Freibauer E., Juvancz I., Káldor I., Kárf J., Kiss A., Klein A., Kohn B., Kornis Ö., Krausz B., Gy. Krisztián, A. Kürth, Gy. Lukhaub, M. Mandel, L. Miliczer, R. Obláth, Gy. Perl, G. Petrogalli, L. Pollák, N. Pollák, J. Porkoláb, K. Pölöskey, Z. Rehberger, D. Róth, G. Sasvári, Ö. Spitzer, S. Szibelth, A. Vida, J. Weisz.