Đáp án:
$S=\left\{k\dfrac{\pi}{2};±\dfrac{1}{2}\arccos\dfrac{1}{8}+k\pi\bigg{|}k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\sin2x-4\sin4x=0$
$⇔\sin2x-8\sin2x\cos2x=0$
$⇔\sin2x(1-8\cos2x)=0$
$⇔\left[ \begin{array}{l}\sin2x=0\\\cos2x=\dfrac{1}{8}\end{array} \right.$
$⇔\left[ \begin{array}{l}2x=k\pi\\2x=±\arccos\dfrac{1}{8}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=k\dfrac{\pi}{2}\\x=±\dfrac{1}{2}\arccos\dfrac{1}{8}+k\pi\end{array} \right.\,\,(k\in\mathbb Z)$
Vậy $S=\left\{k\dfrac{\pi}{2};±\dfrac{1}{2}\arccos\dfrac{1}{8}+k\pi\bigg{|}k\in\mathbb Z\right\}$.
