$\begin{array}{l}
ĐK:\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
{\cos ^2}x + \dfrac{1}{{{{\cos }^2}x}} + \cos x - \dfrac{1}{{\cos x}} - \dfrac{7}{4} = 0\\
\Leftrightarrow {\left( {\cos x - \dfrac{1}{{\cos x}}} \right)^2} + \left( {\cos x - \dfrac{1}{{\cos x}}} \right) + 2 - \dfrac{7}{4} = 0\\
\Leftrightarrow {\left( {\cos x - \dfrac{1}{{\cos x}}} \right)^2} + \left( {\cos x - \dfrac{1}{{\cos x}}} \right) + \dfrac{1}{4} = 0\\
\Leftrightarrow {\left( {\cos x - \dfrac{1}{{\cos x}} + \dfrac{1}{2}} \right)^2} = 0\\
\Leftrightarrow \cos x - \dfrac{1}{{\cos x}} + \dfrac{1}{2} = 0\\
\Leftrightarrow 2{\cos ^2}x - 2 + \cos x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{ - 1 + \sqrt {17} }}{4}\\
\cos x = \dfrac{{ - 1 - \sqrt {17} }}{4}(L)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \arccos \dfrac{{ - 1 + \sqrt {17} }}{4} + k2\pi \\
x = - \arccos \dfrac{{ - 1 + \sqrt {17} }}{4} + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$
