\[\begin{array}{l}
4\sin x + 3\cos x = 4\left( {2 + \tan x} \right) - \frac{5}{{\cos x}}\\
DK:\,\,\,\cos x \ne 0.\\
pt \Leftrightarrow 4\sin x\cos x + 3{\cos ^2}x = 8\cos x + 4\sin x - 5\\
\Leftrightarrow 4\sin x\cos x + 3{\cos ^2}x - 8\cos x - 4\sin x + 5 = 0\\
\Leftrightarrow 4\sin x\left( {\cos x - 1} \right) + \left( {\cos x - 1} \right)\left( {3\cos x - 5} \right) = 0\\
\Leftrightarrow \left( {\cos x - 1} \right)\left( {4\sin x + 3\cos x - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\,\,\,\left( 1 \right)\\
4\sin x + 3\cos x - 5 = 0\,\,\,\,\left( 2 \right)
\end{array} \right.\\
Giai\,\,\,\left( 1 \right) \Leftrightarrow x = k2\pi \,\,\,\left( {tm} \right)\\
\left( 2 \right) \Leftrightarrow \frac{4}{5}\sin x + \frac{3}{5}\cos x = 1\\
\Leftrightarrow \sin \left( {x + \alpha } \right) = 1\,\,\,\left( {\cos \alpha = \frac{4}{5};\,\,\,\sin \alpha = \frac{3}{5}} \right)\\
\Leftrightarrow x + \alpha = \frac{\pi }{2} + k2\pi \\
\Leftrightarrow x = - \alpha + \frac{\pi }{2} + k2\pi \,\,\,\left( {k \in Z} \right).\,
\end{array}\]
