Đáp án:

Giải thích các bước giải:

 `2sin 4x+cos 2x=0`

`⇔ 4sin 2xcos 2x+cos 2x=0`

`⇔ cos 2x(2sin 2x+1)=0`

`⇔` \(\left[ \begin{array}{l}cos 2x=0\\sin 2x=-\dfrac{1}{4}\end{array} \right.\)

\(\begin{array} \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{2} + k\pi\\2x = \arcsin\left(-\dfrac{1}{4}\right) + k2\pi\\2x = \pi - \arcsin\left(-\dfrac{1}{4}\right) + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x = \dfrac{1}{2}\arcsin\left(-\dfrac{1}{4}\right) + k\pi\\x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin\left(-\dfrac{1}{4}\right) + k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}\)

Vậy ..............

Đáp án:

$\left[\begin{array}{l}x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x = \dfrac{1}{2}\arcsin\left(-\dfrac{1}{4}\right) + k\pi\\x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin\left(-\dfrac{1}{4}\right) + k\pi\end{array}\right.\quad (k \in \Bbb Z)$

Giải thích các bước giải:

$\begin{array}{l}2\sin4x + \cos2x = 0\\ \Leftrightarrow 4\sin2x\cos2x + \cos2x = 0\\ \Leftrightarrow \cos2x(4\sin2x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos2x = 0\\\sin2x = - \dfrac{1}{4}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{2} + k\pi\\2x = \arcsin\left(-\dfrac{1}{4}\right) + k2\pi\\2x = \pi - \arcsin\left(-\dfrac{1}{4}\right) + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x = \dfrac{1}{2}\arcsin\left(-\dfrac{1}{4}\right) + k\pi\\x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin\left(-\dfrac{1}{4}\right) + k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$