Đáp án: $x = \pm \frac{\pi }{3} + k2\pi \left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
\frac{{1 + \cos x}}{{\sin x}} = \frac{{\sin 2x}}{{1 - {\mathop{\rm cosx}\nolimits} }}\\
Đkxđ:\left\{ \begin{array}{l}
\sin x \ne 0\\
1 - \cos x \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne k\pi \\
\cos x \ne 1
\end{array} \right. \Rightarrow x \ne k\pi \\
\Rightarrow \left( {1 + \cos x} \right)\left( {1 - \cos x} \right) = \sin x.\sin 2x\\
\Rightarrow 1 - {\cos ^2}x = 2{\sin ^2}x.\cos x\\
\Rightarrow {\sin ^2}x - 2{\sin ^2}x.\cos x = 0\\
\Rightarrow {\sin ^2}x\left( {1 - 2\cos x} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = \frac{1}{2}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = k\pi \left( {loai} \right)\\
x = \pm \frac{\pi }{3} + k2\pi \left( {k \in Z} \right)
\end{array} \right.\\
Vậy\,x = \pm \frac{\pi }{3} + k2\pi \left( {k \in Z} \right)
\end{array}$
