Решите уравнение \(2\cos^2 x + 3\sin 2x = 4 + 3\cos 2x\).

Решение

Воспользуемся формулами синуса и косинуса двойного угла:
\[\hspace{1cm}2\cos^2 x + 6\sin x\cos x = 4 + 3(2\cos^2 x \hspace{0.1cm}- 1)\]\[\hspace{1cm}2\cos^2 x + 6\sin x\cos x = 1 + 6\cos^2 x\]\[\hspace{1cm}2\cos^2 x \hspace{0.2cm}- 1 = 6\cos x (\cos x\hspace{0.2cm} -\hspace{0.1cm} \sin x)\]
Далее, воспользуемся основным тригонометрическим тождеством:
\[\hspace{1cm}2\cos^2 x \hspace{0.2cm}- \hspace{0.1cm}\cos^2 x \hspace{0.2cm}- \hspace{0.1cm}\sin^2 x = 6\cos x (\cos x\hspace{0.2cm} -\hspace{0.1cm} \sin x)\]\[\hspace{1cm}\cos^2 x \hspace{0.2cm}- \hspace{0.1cm}\sin^2 x = 6\cos x (\cos x\hspace{0.2cm} -\hspace{0.1cm} \sin x)\]
И формулой разности квадратов:
\[\hspace{1cm}(\cos x\hspace{0.2cm} -\hspace{0.1cm} \sin x)(\cos x\hspace{0.1cm} +\hspace{0.1cm} \sin x) = 6\cos x (\cos x\hspace{0.2cm} -\hspace{0.1cm} \sin x)\]
Переносим в одну сторону и выделяем общий множитель:
\[\hspace{1cm}(\cos x\hspace{0.2cm} -\hspace{0.1cm} \sin x)(6\cos x\hspace{0.2cm} -\hspace{0.1cm} \cos x\hspace{0.2cm} -\hspace{0.1cm} \sin x) = 0\]\[\hspace{1cm}(\cos x\hspace{0.2cm} -\hspace{0.1cm} \sin x)(5\cos x\hspace{0.2cm} -\hspace{0.1cm} \sin x) = 0\]
Следовательно:
\[\left[
\begin{gathered}
\cos x = \sin x\\
5\cos x = \sin x
\end{gathered}
\right. \Rightarrow
\left[
\begin{gathered}
\mathop{\mathrm{tg}} x = 1\\
\mathop{\mathrm{tg}} x = 5
\end{gathered}
\right.\]

Откуда:
\[\left[
\begin{gathered}
x = \frac{\pi}{4} + \pi n\\
x = \mathop{\mathrm{arctg}} 5 + \pi n
\end{gathered}
\right.,n \in \mathbb{Z}.\]

Ответ: \(\frac{\pi}{4} + \pi n,\mathop{\mathrm{arctg}} 5 + \pi n, \ n \in \mathbb{Z}\).