Đáp án:
\(\begin{array}{l}
1,\\
x = - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\\
2,\\
x = - \dfrac{\pi }{2} + k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
4\sin x.\cos x.\cos 2x = - 1\\
\Leftrightarrow 2.\left( {2\sin x.\cos x} \right).\cos 2x = - 1\\
\Leftrightarrow 2.\sin 2x.\cos 2x = - 1\\
\Leftrightarrow \sin \left( {2.2x} \right) = - 1\\
\Leftrightarrow \sin 4x = - 1\\
\Leftrightarrow 4x = - \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\\
2,\\
DKXD:\,\,\,1 - \cos 2x \ne 0 \Leftrightarrow \cos 2x \ne 1\\
\dfrac{{2\sin 2x}}{{1 - \cos 2x}} = 0\\
\Leftrightarrow 2\sin 2x = 0\\
\Leftrightarrow \sin 2x = 0\\
{\sin ^2}2x + {\cos ^2}2x = 1\\
\Leftrightarrow {\cos ^2}2x = 1\\
\cos 2x \ne 1 \Rightarrow \cos 2x = - 1\\
\Leftrightarrow 2x = - \pi + k2\pi \\
\Leftrightarrow x = - \dfrac{\pi }{2} + k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
