Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinxcosx=t^2-1\end{matrix}\right.\)
Pt trở thành:
\(\left(1+\sqrt{2}\right)t-t^2+1-1-\sqrt{2}=0\)
\(\Leftrightarrow t^2-\left(1+\sqrt{2}\right)t+\sqrt{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow x=?\)
