8cos2x.sin2x.cos4x=cân 2

Đáp án:

Giải thích các bước giải: $\begin{array}{l}8cos2x.sin2x.cos4x = \sqrt 2 {\rm{ }}\\ \Leftrightarrow {\rm{4}}{\rm{.2}}{\rm{.}}\,cos2x.\sin 2x.cos4x = \sqrt 2 \\ \Leftrightarrow 4.\,\,\sin \,4x.cos4x = \sqrt 2 \\ \Leftrightarrow 2.2\,\sin \,4x.cos4x = \sqrt 2 \\ \Leftrightarrow 2.\sin 8x = \sqrt 2 \\ \Leftrightarrow \sin 8x = \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \sin 8x = \sin \frac{\pi }{4}\\ \Leftrightarrow \left\{ \begin{array}{l}8x = \frac{\pi }{4} + k2\pi ,\,\,\,k \in Z\\8x = \pi - \frac{\pi }{4} + k2\pi ,\,k \in Z\,\,\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = \frac{\pi }{{32}} + \frac{{k\pi }}{4},\,\,k \in Z\\8x = \frac{{3\pi }}{4} + k2\pi ,\,k \in Z\,\,\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = \frac{\pi }{{32}} + \frac{{k\pi }}{4},\,\,k \in Z\\x = \frac{{3\pi }}{{32}} + \frac{{k\pi }}{4},\,k \in Z\,\,\end{array} \right.\end{array}$

Đáp án:

$\left\{ \begin{array}{l} x = \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}\\ x = \dfrac{{3\pi }}{{32}} + \dfrac{{k\pi }}{4} \end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right).$

Lời giải:

$\begin{array}{l} 8\cos 2x.\sin 2x.\cos 4x = \sqrt 2 \\ \Leftrightarrow 4\sin 4x.\cos 4x = \sqrt 2 \\ \Leftrightarrow 2\sin 8x = \sqrt 2 \\ \Leftrightarrow \sin 8x = \dfrac{{\sqrt 2 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l} 8x = \dfrac{\pi }{4} + k2\pi \\ 8x = \dfrac{{3\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}\\ x = \dfrac{{3\pi }}{{32}} + \dfrac{{k\pi }}{4} \end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right). \end{array}$

Vậy $\left\{ \begin{array}{l} x = \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}\\ x = \dfrac{{3\pi }}{{32}} + \dfrac{{k\pi }}{4} \end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right).$