Lời giải:
Ta có:
\(y=\frac{1+\cos x}{3-\sin x}\Rightarrow y'=\frac{-\sin x(3-\sin x)-(1+\cos x)-\cos x}{(3-\sin x)^2}\)
\(=\frac{\sin ^2x+\cos^2x-3\sin x+\cos x}{(3-\sin x)^2}=\frac{1-3\sin x+\cos x}{(3-\sin x)^2}\)
\(y'=0\Leftrightarrow 3\sin x-\cos x=1\)
Kết hợp \(\sin ^2x+\cos ^2x=1\)
\(\Rightarrow \sin ^2x+(3\sin x-1)^2=1\)
\(\Rightarrow \left[\begin{matrix} \sin x=0\rightarrow \cos x=1\\ \sin x=\frac{3}{5}\rightarrow \cos x=\frac{4}{5}\end{matrix}\right.\)
Vậy \(y'=0\Leftrightarrow (\cos x, \sin x)=(1;0)\) hoặc \((\frac{4}{5};\frac{3}{5})\)
Thay thế giá trị trên ta thấy \(y_{\max}=\frac{3}{4}\Leftrightarrow (\cos x,\sin x)=(\frac{4}{5}; \frac{3}{5})\)
