Đk:\(\frac{\pi}{4}+\frac{k\pi}{2}\le x\le\frac{k\pi}{2}\)
\(2sin2x+\sqrt{2}sin4x=0\)
\(\Rightarrow2sin2x+\sqrt{2}\cdot2sin2x\cdot cos2x=0\)
\(\Rightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Rightarrow sin2x=0\)hoặc \(1+\sqrt{2}cos2x=0\)
Xét \(sin2x=0\Rightarrow x=\frac{k\pi}{2}\)\(\left(k\in Z\right)\)Xét \(1+\sqrt{2}cos2x=0\)\(\Rightarrow cos2x=\frac{-1}{\sqrt{2}}\)
\(\Rightarrow2x=\pm\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\pm\frac{\pi}{8}+k\pi\left(k\in Z\right)\)
Vậy
Ta có: \(2sin2x+\sqrt{2}sin4x=0\)
\(\Rightarrow2sin2x+2\sqrt{2}sin2xcos2x=0\)
\(\Rightarrow sin2x+\sqrt{2}sin2xcos2x=0\)
\(\Rightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}sin2x=0\left(1\right)\\cos2x=-\frac{1}{\sqrt{2}}\left(2\right)\end{array}\right.\)
\(\left(1\right)\Rightarrow2x=k\pi\Rightarrow x=\frac{k\pi}{2}\)
\(\left(2\right)\Rightarrow cos2x=cos\frac{3\pi}{4}\Rightarrow\left[\begin{array}{nghiempt}2x=\frac{3\pi}{4}+k2\pi\left(3\right)\\2x=-\frac{3\pi}{4}+k2\pi\left(4\right)\end{array}\right.\)
\(\left(3\right)\Rightarrow x=\frac{3\pi}{8}+k\pi\)
\(\left(4\right)\Rightarrow x=-\frac{3\pi}{8}+k\pi\)
Vậy \(x=\left\{\frac{k\pi}{2};\frac{3\pi}{8}+k\pi;-\frac{3\pi}{8}+k\pi\right\}\)
\(2sin2x+\sqrt{2}sin4x=0\) đề thế này à
