Ta có: \(8{\sin}^6 x={\sin}^2 2x\)
\(\Leftrightarrow 8{\sin}^6 x=4{\sin}^2 x{\cos}^2 x\)
\(\Leftrightarrow 4{\sin}^2 x(2{\sin}^4 x+{\sin}^2 x-1)=0\)
\( \Leftrightarrow \left[ \begin{array}{l}{\sin}^2 x = 0\\2{\sin}^4 x+{\sin}^2 x-1=0\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l} x = k\pi,k\in\mathbb{Z}\\{\sin}^2 x=\dfrac{1}{2}\\{\sin}^2 x=-1\le 0\text{(loại)}\end{array} \right.\)
Với: \({\sin}^2 x=\dfrac{1}{2}\)
\(\Leftrightarrow \dfrac{1-\cos 2x}{2}=\dfrac{1}{2}\)
\(\Leftrightarrow \cos 2x=0\)
\(\Leftrightarrow 2x=\dfrac{\pi}{2}+k\pi,k\in\mathbb{Z}\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\dfrac{\pi}{2},k\in\mathbb{Z}\)
Vậy phương trình có nghiệm là \(x=k\pi,k\in\mathbb{Z}\) và \(x=\dfrac{\pi}{4}+k\dfrac{\pi}{2},k\in\mathbb{Z}\)
Đáp án: D.
