Giải phương trình: cot² x – tan² x = 16cos 2x.

Ta có: \({\cot ^2}x - {\tan ^2}x = \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\)

\( = \frac{{{{\cos }^4}x - {{\sin }^4}x}}{{{{\sin }^2}x\,.\,{{\cos }^2}x}} = \frac{{4\cos 2x}}{{{{\sin }^2}2x}}\).

Điều kiện: \(\left\{ \begin{array}{l}\sin 2x \ne 0\\\cos 2x \ne 0\end{array} \right. \Leftrightarrow \sin 4x \ne 0\).

Lúc đó: cot² x – tan² x = 16cos 2x

\( \Leftrightarrow \frac{{4\cos 2x}}{{{{\sin }^2}2x}} = 16\cos 2x\)

\( \Leftrightarrow \cos 2x\left( {\frac{1}{{{{\sin }^2}2x}} - 4} \right) = 0\)

+) TH1: cos 2x = 0

\( \Rightarrow 2x = \frac{\pi }{2} + k\pi ,\;k \in \mathbb{Z}\)

\( \Rightarrow x = \frac{\pi }{4} + k\frac{\pi }{2},\;k \in \mathbb{Z}\) (thỏa mãn)

+) TH2: \(\frac{1}{{{{\sin }^2}2x}} - 4 = 0\)

\( \Leftrightarrow {\sin ^2}2x = \frac{1}{4}\)

\( \Leftrightarrow \sin 2x = \pm \frac{1}{2}\)

• Với \(\sin 2x = \frac{1}{2}\), ta có:

\(\left[ \begin{array}{l}2x = \frac{\pi }{6} + k2\pi \;\left( {k \in \mathbb{Z}} \right)\\2x = \frac{{5\pi }}{6} + k2\pi \;\left( {k \in \mathbb{Z}} \right)\end{array} \right. \Rightarrow \left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \;\left( {k \in \mathbb{Z}} \right)\\x = \frac{{5\pi }}{{12}} + k\pi \;\left( {k \in \mathbb{Z}} \right)\end{array} \right.\) (thỏa mãn)

- Với \(\sin 2x = - \frac{1}{2}\), ta có:

\(\left[ \begin{array}{l}2x = - \frac{\pi }{6} + k2\pi \;\left( {k \in \mathbb{Z}} \right)\\2x = \frac{{7\pi }}{6} + k2\pi \;\left( {k \in \mathbb{Z}} \right)\end{array} \right. \Rightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k\pi \;\left( {k \in \mathbb{Z}} \right)\\x = \frac{{7\pi }}{{12}} + k\pi \;\left( {k \in \mathbb{Z}} \right)\end{array} \right.\) (thỏa mãn)

Vậy các họ nghiệm của phương trình là:

\(x \in \left\{ {\frac{\pi }{4} + k\frac{\pi }{2};\; \pm \frac{\pi }{{12}} + k\pi ;\;\frac{{5\pi }}{{12}} + k\pi ;\;\frac{{7\pi }}{{12}} + k\pi \;\left( {k \in \mathbb{Z}} \right)} \right\}\)