Đáp án:
\(\begin{array}{l}
2,\\
\left[ \begin{array}{l}
x = k\pi \\
x = \arccos \dfrac{{ - 1}}{6} + k2\pi \\
x = - \arccos \dfrac{{ - 1}}{6} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\left[ \begin{array}{l}
x = \dfrac{{ - \pi }}{{24}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{7\pi }}{{24}} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)\\
\sin 2x = 2\sin x.\cos x\\
2,\\
\cos \left( {\dfrac{\pi }{2} - x} \right) + 3\sin 2x = 0\\
\Leftrightarrow \sin x + 3.2\sin x.\cos x = 0\\
\Leftrightarrow \sin x + 6\sin x.\cos x = 0\\
\Leftrightarrow \sin x.\left( {1 + 6\cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
1 + 6\cos x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = - \dfrac{1}{6}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \arccos \dfrac{{ - 1}}{6} + k2\pi \\
x = - \arccos \dfrac{{ - 1}}{6} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
8\sin x.\cos x.\cos 2x + 1 = 0\\
\Leftrightarrow 4.\left( {2\sin x.\cos x} \right).\cos 2x + 1 = 0\\
\Leftrightarrow 4\sin 2x.\cos 2x + 1 = 0\\
\Leftrightarrow 2.\left( {2\sin 2x.\cos 2x} \right) + 1 = 0\\
\Leftrightarrow 2.\sin \left( {2.2x} \right) + 1 = 0\\
\Leftrightarrow 2\sin 4x + 1 = 0\\
\Leftrightarrow \sin 4x = - \dfrac{1}{2}\\
\Leftrightarrow \sin 4x = \sin \left( { - \dfrac{\pi }{6}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
4x = - \dfrac{\pi }{6} + k2\pi \\
4x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - \pi }}{{24}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{7\pi }}{{24}} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
