Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
*)\\
{\sin ^3}\left( {x + \dfrac{{2\pi }}{3}} \right) = - \dfrac{1}{8}\\
\Leftrightarrow \sin \left( {x + \dfrac{{2\pi }}{3}} \right) = - \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{{2\pi }}{3} = - \dfrac{\pi }{6} + k2\pi \\
x + \dfrac{{2\pi }}{3} = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{{5\pi }}{6} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
*)\\
{\sin ^4}x = \dfrac{9}{{16}}\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = \dfrac{{\sqrt 3 }}{2}\\
\sin x = - \dfrac{{\sqrt 3 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{\pi }{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
*)\\
DKXD:\,\,\,1 + \cos x \ne 0 \Leftrightarrow \cos x \ne - 1 \Leftrightarrow x \ne \pi + k2\pi \\
\dfrac{{\sin x}}{{1 + \cos x}} = 0\\
\Leftrightarrow \sin x = 0\\
\Leftrightarrow x = k\pi \\
x \ne \pi + k2\pi \Rightarrow x = k2\pi
\end{array}\)
