Giải pt : 4√3 . cosx . sinx . cos2x = sin8x ? 27/09/2021 Bởi aihong Giải pt : 4√3 . cosx . sinx . cos2x = sin8x ?
Đáp án: \[\left[ \begin{array}{l} x = \frac{{k\pi }}{4}\\ x = \frac{\pi }{{24}} + \frac{{k\pi }}{2}\\ x = – \frac{\pi }{{24}} + \frac{{k\pi }}{2} \end{array} \right.\,\,\,\left( {k \in Z} \right)\] Giải thích các bước giải: \[\begin{array}{l} 4\sqrt 3 \cos x.\sin x.\cos 2x = \sin 8x\\ \Leftrightarrow 2\sqrt 3 \sin 2x\cos 2x = 2\sin 4x\cos 4x\\ \Leftrightarrow \sqrt 3 \sin 4x – 2\sin 4x\cos 4x = 0\\ \Leftrightarrow \sin 4x\left( {\sqrt 3 – 2\cos 4x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin 4x = 0\\ \cos 4x = \frac{{\sqrt 3 }}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 4x = k\pi \\ 4x = \frac{\pi }{6} + k2\pi \\ 4x = – \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{{k\pi }}{4}\\ x = \frac{\pi }{{24}} + \frac{{k\pi }}{2}\\ x = – \frac{\pi }{{24}} + \frac{{k\pi }}{2} \end{array} \right.\,\,\,\left( {k \in Z} \right). \end{array}\] Bình luận
Đáp án:
\[\left[ \begin{array}{l}
x = \frac{{k\pi }}{4}\\
x = \frac{\pi }{{24}} + \frac{{k\pi }}{2}\\
x = – \frac{\pi }{{24}} + \frac{{k\pi }}{2}
\end{array} \right.\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
\[\begin{array}{l}
4\sqrt 3 \cos x.\sin x.\cos 2x = \sin 8x\\
\Leftrightarrow 2\sqrt 3 \sin 2x\cos 2x = 2\sin 4x\cos 4x\\
\Leftrightarrow \sqrt 3 \sin 4x – 2\sin 4x\cos 4x = 0\\
\Leftrightarrow \sin 4x\left( {\sqrt 3 – 2\cos 4x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 4x = 0\\
\cos 4x = \frac{{\sqrt 3 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4x = k\pi \\
4x = \frac{\pi }{6} + k2\pi \\
4x = – \frac{\pi }{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{k\pi }}{4}\\
x = \frac{\pi }{{24}} + \frac{{k\pi }}{2}\\
x = – \frac{\pi }{{24}} + \frac{{k\pi }}{2}
\end{array} \right.\,\,\,\left( {k \in Z} \right).
\end{array}\]