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