Đặt \(cosx-sinx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(2t+1-t^2+2=0\)
\(\Leftrightarrow-t^2+2t+3=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cosx-sinx=-1\)
\(\Leftrightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
Lời giải:
PT $\Leftrightarrow 2(\cos x-\sin x)+2\sin x\cos x+2=0$
$\Leftrightarrow \cos x-\sin x+\sin x\cos x+1=0$
$\Leftrightarrow (\cos x+1)+\sin x(\cos x-1)=0$
$\Leftrightarrow 2\cos ^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}(\cos x-1)=0$
$\Leftrightarrow 2\cos \frac{x}{2}[\cos \frac{x}{2}+\sin \frac{x}{2}(\cos x-1)]=0$
Nếu $\cos \frac{x}{2}=0\Rightarrow x=2k\pi +\pi $ với $k$ nguyên.
Nếu $\cos \frac{x}{2}+\sin \frac{x}{2}(\cos x-1)=0$
$\Leftrightarrow (\cos \frac{x}{2}-\sin \frac{x}{2})+\sin \frac{x}{2}(\cos \frac{x}{2}-\sin \frac{x}{2})(\cos \frac{x}{2}+\sin \frac{x}{2})=0$
$\Leftrightarrow (\cos \frac{x}{2}-\sin \frac{x}{2})(1+\sin ^2\frac{x}{2}+\sin \frac{x}{2}.\cos \frac{x}{2})=0$
Dễ thấy $1+\sin ^2\frac{x}{2}+\sin \frac{x}{2}.\cos \frac{x}{2}>0$ nên $\cos \frac{x}{2}-\sin \frac{x}{2}=0$
$\Rightarrow \cos \frac{x}{2}=\sin \frac{x}{2}=\frac{\pm 1}{\sqrt{2}}$
$\Rightarrow x=2k\pi +\frac{\pi}{2}$ với $k$ nguyên.
