b)
\(\begin{array}{l}
\dfrac{{\sqrt 3 }}{{\cos x}} + \dfrac{1}{{\sin x}} = 8\sin x\,\,DK:\sin 2x \ne 0\\
\Rightarrow \sqrt 3 \sin x + \cos x = 8{\sin ^2}x\cos x\\
\Leftrightarrow \sqrt 3 .\dfrac{{\sin x}}{{{{\sin }^3}x}} + \dfrac{{\cos x}}{{{{\sin }^3}x}} = 8.\dfrac{{{{\sin }^2}x\cos x}}{{{{\sin }^3}x}}\\
\Leftrightarrow \sqrt 3 .\frac{1}{{{{\sin }^2}x}} + \cot x.\frac{1}{{{{\sin }^2}x}} = 8.\cot x\\
\Leftrightarrow \sqrt 3 \left( {1 + {{\cot }^2}x} \right) + \cot x\left( {1 + {{\cot }^2}x} \right) = 8\cot x\\
\Leftrightarrow {\cot ^3}x + \sqrt 3 {\cot ^2}x - 7\cot x + \sqrt 3 = 0\\
\Leftrightarrow \left( {\cot x - \sqrt 3 } \right)\left( {{{\cot }^2}x + 2\sqrt 3 \cot x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cot x = \sqrt 3 \\
\cot x = 2 - \sqrt 3 \\
\cot x = 2 + \sqrt 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k\pi \\
x = {\mathop{\rm arccot}\nolimits} \left( {2 - \sqrt 3 } \right) + k\pi \\
x = {\mathop{\rm arccot}\nolimits} \left( {2 + \sqrt 3 } \right) + k\pi
\end{array} \right.
\end{array}\)
