\(\frac{1}{2}+\frac{1}{2}cos2x+\sqrt{2}sin2x+1=0\)
\(\Leftrightarrow cos2x+2\sqrt{2}sin2x=-3\)
\(\Leftrightarrow\frac{2\sqrt{2}}{3}sin2x+\frac{1}{3}cos2x=-1\)
Đặt \(\frac{2\sqrt{2}}{3}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow sin2x.cosa+cos2x.sina=-1\)
\(\Leftrightarrow sin\left(2x+a\right)=-1\)
\(\Leftrightarrow2x+a=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{a}{2}-\frac{\pi}{4}+k\pi\)
\(cos^2x+\sqrt{2}sin2x+1=0\)
\(\Leftrightarrow cos^2x+\sqrt{2}.2sinxcosx+cos^2x+sin^2x=0\)
\(\Leftrightarrow2cosx^2+\sqrt{2}.2cosxsinx+sin^2x=0\)
\(\Leftrightarrow\left(\sqrt{2}cosx+sinx\right)^2=0\)
\(\Rightarrow\sqrt{2}cosx+sinx=0\)
\(\Leftrightarrow sinx=-\sqrt{2}cosx\)
\(\Rightarrow tanx=-\sqrt{2}\left(x\ne\frac{\pi}{2}+k\pi\right)\)
\(\Leftrightarrow x=arctan\left(-\sqrt{2}\right)+k\pi\)
