\(12{\rm{arct}}{{\rm{g}}^2}\dfrac{x}{2} = \pi \left( {3\pi + 5{\rm{arctg}}\dfrac{x}{2}} \right).\)
Пусть \({\rm{arctg}}\dfrac{x}{2} = t,\) где \(\,t\, \in \,\left( {-\dfrac{\pi }{2};\dfrac{\pi }{2}} \right).\) Тогда:
\(12{t^2}-5\pi t-3{\pi ^2} = 0;\,\,\,\,D = 25{\pi ^2} + 144{\pi ^2} = 169{\pi ^2};\,\,\,\,\,\left[ \begin{array}{l}t = \dfrac{{5\pi + 13\pi }}{{24}} = \dfrac{{3\pi }}{4}\,\, \notin \,\left( {-\dfrac{\pi }{2};\dfrac{\pi }{2}} \right).\\t = \dfrac{{5\pi -13\pi }}{{24}} = -\dfrac{\pi }{3}.\end{array} \right.\)
Возвращаясь к прежней переменной, получим:
\({\rm{arctg}}\dfrac{x}{2} = -\dfrac{\pi }{3}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{x}{2} = -\sqrt 3 \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -2\sqrt 3 .\)
Ответ: \(-2\sqrt 3 \).