Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le2\\sin2x=2sinx.cosx=t^2-1\end{matrix}\right.\)
Pt trở thành:
\(t+t^2-1=1+\sqrt{2}\)
\(\Leftrightarrow t^2+t-2-\sqrt{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\sqrt{2}\\t=-1-\sqrt{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow...\)
Đúng 0
Bình luận (0)
