Ai giải cho mình bài này với ạ xíu mình nộp bài tết 8 \cos x=\frac{\sqrt{3}}{\sin x}+\frac{1}{\cos x}

Điều kiện xác định:

$\begin{array}{l} \left\{ \begin{array}{l} \sin x \ne 0\\ \cos x \ne 0 \end{array} \right. \Rightarrow \sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi \\  \Leftrightarrow x \ne \dfrac{{k\pi }}{2}\left( {k \in \mathbb Z} \right) \end{array}$

$\begin{array}{l}
8\cos x = \dfrac{{\sqrt 3 }}{{\sin x}} + \dfrac{1}{{\cos x}}\\
 \Leftrightarrow 8\cos x = \dfrac{{\sqrt 3 \cos x + \sin x}}{{\sin x\cos x}}\\
 \Leftrightarrow 8{\cos ^2}x\sin x = \sqrt 3 \cos x + \sin x\\
 \Leftrightarrow 2\cos x\sin x.4\cos x = \sqrt 3 \cos x + \sin x\\
 \Leftrightarrow 4\cos x\sin 2x = \sqrt 3 \cos x + \sin x\\
 \Leftrightarrow 2\sin 3x + 2\sin x = \sqrt 3 \cos x + \sin x\\
 \Leftrightarrow 2\sin 3x = \sqrt 3 \cos x - \sin x\\
 \Leftrightarrow 2\sin 3x = 2\cos \left( {x + \dfrac{\pi }{3}} \right) \Leftrightarrow \cos \left( {\dfrac{\pi }{2} - 3x} \right) = \cos \left( {x + \dfrac{\pi }{3}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{2} - 3x = x + \dfrac{\pi }{3} + k2\pi \\
\dfrac{\pi }{2} - 3x =  - x - \dfrac{\pi }{3} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{6} - k2\pi \\
2x = \dfrac{{5\pi }}{6} - k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} - k\dfrac{\pi }{2}\\
x = \dfrac{{5\pi }}{{12}} - k\pi 
\end{array} \right.\left( {k \in \mathbb Z} \right)
\end{array}$