\(-1\le sinx;cosx\le1\Rightarrow\left\{{}\begin{matrix}0\le1+sinx\le2\\1\le2+cosx\le3\end{matrix}\right.\)
\(\Rightarrow0\le y\le2\)
Th1: \(y=0\Rightarrow sinx=-1\Rightarrow x=-\frac{\pi}{2}+k2\pi\)
Th2: \(y=1\Rightarrow1+sinx=2+cosx\)
\(\Rightarrow sinx-cosx=1\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Rightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
TH3: \(y=2\Rightarrow1+sinx=4+2cosx\)
\(\Leftrightarrow sinx=3+2cosx\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sinx=1\\cosx=-1\end{matrix}\right.\) (vô nghiệm)
