Решите уравнение \(\mathop{\mathrm{tg}} 2x = 2\cos 2x \mathop{\mathrm{ctg}} x.\)

Решение

\[
\mathop{\mathrm{tg}} 2x = 2\cos 2x \mathop{\mathrm{ctg}} x \quad\Leftrightarrow
\] \[
\dfrac{\sin 2x}{\cos 2x} = 2\cos 2x \cdot\dfrac{\cos x}{\sin x} \quad\Leftrightarrow
\] \[
\dfrac{2\sin x \cos x}{\cos 2x} = 2\cos 2x \cdot\dfrac{\cos x}{\sin x} \quad\Leftrightarrow
\] \[
\left\{\begin{array}{l} \sin x \cos 2x \ne 0 \\ \sin x \sin 2x = 2\cos^2 2x \cos x \end{array}\right. \quad\Leftrightarrow
\] \[
\left\{\begin{array}{l} \sin x \cos 2x \ne 0 \\ 2 \sin x \sin x \cos x = 2\cos^2 2x \cos x \end{array}\right. \quad\Leftrightarrow
\] \[
\left\{\begin{array}{l} \sin x \cos 2x \ne 0 \\ \cos x (2\cos^2 2x - 2\sin^2 x) = 0 \end{array}\right. \quad\Leftrightarrow
\] \[
\left\{\begin{array}{l} \sin x \cos 2x \ne 0 \\ \cos x (2\cos^2 2x + \cos 2x -1) = 0 \end{array}\right. \quad\Leftrightarrow\quad
\] \[ \left\{\begin{array}{l} \sin x \cos 2x \ne 0 \\ \cos x (2\cos 2x - 1)(\cos 2x + 1) = 0 \end{array}\right. \quad\Leftrightarrow
\] \[
\left\{\begin{array}{l} \sin x \cos 2x \ne 0 \\ \left[\begin{array}\\ \cos x = 0 \\ \cos 2x = \dfrac{1}{2}\\ \cos 2x = -1 \end{array}\right. \end{array}\right. \quad\Leftrightarrow
\] \[
\left[\begin{array}{l} x = \pm\dfrac{\pi}{6} + \pi n, \quad n \in \mathbb Z \\ \\ x = \dfrac{\pi}{2} + \pi k, \quad k \in \mathbb Z\end{array}\right. \]

Ответ: \(\pm\dfrac{\pi}{6} + \pi n, \dfrac{\pi}{2} + \pi k \mid n, k \in \mathbb Z\).