\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)+1=3sinx.cosx\)
Đặt \(sinx+cosx=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)
Phương trình trở thành:
\(t\left(1-\dfrac{t^2-1}{2}\right)+1=\dfrac{3}{2}\left(t^2-1\right)\)
\(\Leftrightarrow t^3+3t^2-3t-5=0\)
\(\Leftrightarrow\left(t+1\right)\left(t^2+2t-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=-1-\sqrt{6}\left(loại\right)\\t=-1+\sqrt{6}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
