Giải phương trình: \({\tan ^2}x - \dfrac{4}{{\cos x}} + 5 = 0\).
Giải chi tiết:
\({\tan ^2}x - \dfrac{4}{{\cos x}} + 5 = 0\) (ĐK: \(x \ne \dfrac{\pi }{2} + k\pi \)).
\(\begin{array}{l} \Leftrightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - \dfrac{4}{{\cos }} + 5 = 0 \Leftrightarrow {\sin ^2}x + 5{\cos ^2}x - 4\cos x = 0\\ \Leftrightarrow \left( {1 - {{\cos }^2}x} \right) + 5{\cos ^2}x - 4\cos x = 0 \Leftrightarrow 4{\cos ^2}x - 4\cos x + 1 = 0\\ \Leftrightarrow \cos x = \dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{3} + k2\pi \\x = - \dfrac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)