Giải phương trình \(2\sin2x+\sqrt{2}\sin4x=0\).
\(\left[\begin{array}{nghiempt}x=k\frac{\pi}{2}\\x=\pm\frac{3\pi}{8}+k\pi\end{array}\right.\)với \(k\in Z.\)\(x=2k\pi\) với \(k\in Z.\)\(x=k\dfrac{\pi}{2}\) với \(k\in Z.\)\(x=\pm\dfrac{3\pi}{8}+2k\pi\) với \(k\in Z.\)Hướng dẫn giải:\(2\sin2x+\sqrt{2}\sin4x=0\)
\(\Leftrightarrow2\sin2x+\sqrt{2}\left(2\sin2x.\cos2x\right)=0\)
\(\Leftrightarrow2\sin2x\left(1+\sqrt{2}\cos2x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sin2x=0\\\cos2x=-\frac{1}{\sqrt{2}}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=k\pi\\2x=\pm\frac{3\pi}{4}+2k\pi\end{array}\right.\) (\(k\in\mathbb{Z}\))
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=k\frac{\pi}{2}\\x=\pm\frac{3\pi}{8}+k\pi\end{array}\right.\)