Đk: $\left\{ \begin{array}{l} {\cos }^2x\ne0\\ \sin x\ne0\\\cos2x\ne0\end{array} \right .$
Phương trình tương đương:
\(\begin{array}{l}
\dfrac{{3\sin x - 4{{\sin }^3}x}}{{\sin x}} - \dfrac{2}{{2{{\cos }^2}x - 1}} = 1 + {\tan ^2}x\\
\Rightarrow 3 - 4{\sin ^2}x - \dfrac{2}{{2{{\cos }^2}x - 1}} = 1 + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\\
\Rightarrow 3 - 4(1 - co{{\mathop{\rm s}\nolimits} ^2}x) - \dfrac{2}{{2{{\cos }^2}x - 1}} = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}x}}\\
\Rightarrow ( - 1 + 4t)(2t - 1)t - 2t = 2t - 1\\
\Rightarrow 8{t^3} - 6{t^2} - 3t + 1 = 0\\
\Rightarrow \left[ \begin{array}{l}
t = 1\\
t = \dfrac{{ - 1}}{2} < 0(l)\\
t = \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{\cos ^2}x = 1(l)\\
{\cos ^2}x = \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \cos x = \dfrac{{ \pm 1}}{2}
\end{array}\)
