Выполним следующее преобразование: \(6 = 6 \cdot 1 = 6\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) = 6{\sin ^2}\alpha + 6{\cos ^2}\alpha \)
Тогда:
\(4{\sin ^2}\alpha + 9{\cos ^2}\alpha = 6\,\,\,\,\, \Leftrightarrow \,\,\,\,\,4{\sin ^2}\alpha + 9{\cos ^2}\alpha = 6{\sin ^2}\alpha + 6{\cos ^2}\alpha \,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,2{\sin ^2}\alpha = 3{\cos ^2}\alpha \,\,\,\,\, \Leftrightarrow \,\,\,\,\,\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}\, = \frac{3}{2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,t{g^2}\alpha = 1,5.\)
Ответ: 1,5.