XLVII OM - I - Problem 1

Determine all integers $ n $ for which the equation $ 2 \sin nx = \tan x + \cot x $ has solutions in real numbers $ x $.
Assume that the given equation has solutions in real numbers: let $ \alpha $ be one of these solutions. Of course, $ \alpha $ cannot be an integer multiple of $ \pi/2 $, because for such numbers the right side loses its meaning. According to the equation,






hence $ 2\sin n\alpha \cdot \sin 2\alpha=2 $, which means






(we use the formula: $ \cos u - \cos v = 2\sin \frac{1}{2}(u+v)\sin\frac{1}{2}(v-u)) $. The difference of two cosines is equal to $ 2 $ if and only if the first one has the value $ 1 $ and the second one $ -1 $. We thus obtain the system of equalities






from which it follows that






($ p, q $ are integers).
In the second of the equalities (1), we have a non-zero number on the right side; hence the left side also has a non-zero value. We can therefore divide the first equality by the second. We get the relation $ \frac{n-2}{n+2}=\frac{2q}{2p+1} $.
By introducing the notation $ n-2 = m $, we rewrite this relation as $ (2p+1)m = 2q(m+4) $, or






We see that the number $ m $ multiplied by an odd number gives a result divisible by $ 8 $, which is possible only if the number $ m $ itself is divisible by $ 8 $. Since $ n = m + 2 $, this means that the number $ n $ gives a remainder of $ 2 $ when divided by $ 8 $.
Taking into account the condition of the problem that the number $ n $ must be positive, we arrive at the following conclusion:






Conversely, if the number $ n $ has such a form, then the considered equation has solutions in real numbers --- one of them is, for example, the number $ x = \frac{1}{4} \pi $; indeed: for $ x=\frac{1}{4} \pi $ and $ n = 8k + 2 $ we have $ nx = 2k\pi + \frac{1}{2} \pi $, from which $ \sin nx = 1 $; both the left and the right side of the equation have the value $ 2 $.
Conclusion: the numbers whose determination was the goal of the problem are characterized by the statement (2).
Note: We will also mention a variant of the solution based on the observation (right at the beginning) that for any number $ x $ the right side of the equation, as a number of the form $ t + (1/t) $, has an absolute value not less than $ 2 $, while the absolute value of the left side is not greater than $ 2 $. The equation can therefore be satisfied only when both sides have the value $ 2 $ or $ -2 $. Each of these possibilities gives a system of two simple trigonometric equations. Completing the reasoning is an easy task and leads to conclusion (2), which characterizes the sought numbers $ n $. We leave the elaboration of the details to the reader.