\[\begin{array}{l}
{\text{Решить уравнение:}} \hfill \\
2{\sin ^3}x = \cos x. \hfill \\
\end{array}\]
\[2{\sin ^3}x = \cos x \cdot \left( {{{\sin }^2}x + {{\cos }^2}x} \right) \Leftrightarrow 2{\operatorname{tg} ^3}x - {\operatorname{tg} ^2}x - 1 = 0\]
\[\frac{\pi }{4} + \pi k\]
