Đáp án:
\(\left[ \begin{array}{l}
x = k\pi \\
x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:\cos x \ne - 1 \to x \ne \pi + k2\pi \\
\dfrac{{\sin x}}{{\cos x + 1}} = 1 - \cos x\\
\to \sin x = \left( {1 - \cos x} \right)\left( {1 + \cos x} \right)\\
\to \sin x = 1 - {\cos ^2}x\\
\to \sin x = {\sin ^2}x\\
\to \sin x\left( {\sin x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
\sin x = 0\\
\sin x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = k\pi \\
x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
