Phương pháp giải:

Đặt \(\sin x=a,\,\,\cos x=b\)

Lời giải chi tiết:

Đặt  ta có \({{a}^{2}}+{{b}^{2}}=1\).

Khi  đó \(y=\left| a+b+\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b} \right|=\left| \frac{ab\left( a+b \right)+{{a}^{2}}+{{b}^{2}}+a+b}{ab} \right|=\left| \frac{ab\left( a+b \right)+a+b+1}{ab} \right|\)

Đặt \(t=a+b\in \left[ -\sqrt{2};\sqrt{2} \right]\Rightarrow {{t}^{2}}={{a}^{2}}+{{b}^{2}}+2ab=1+2ab\Rightarrow ab=\frac{{{t}^{2}}-1}{2}\) , khi đó ta có :

\(y=\left| t+\frac{2\left( t+1 \right)}{{{t}^{2}}-1} \right|=\left| t+\frac{2}{t-1} \right|=\left| t-1+\frac{2}{t-1}+1 \right|\)

Nếu \(t-1>0\Rightarrow t-1+\frac{2}{t-1}+1\ge 2\sqrt{2}+1\Rightarrow y\ge 2\sqrt{2}+1\)

Nếu \(t-1<0\Rightarrow \frac{1}{1-t}+1-t\ge 2\sqrt{2}\Rightarrow \frac{1}{t-1}+t-1\le -2\sqrt{2}\Rightarrow \frac{1}{t-1}+t-1+1\le 1-2\sqrt{2}\Rightarrow y\ge 2\sqrt{2}-1\)

Vậy \(y\ge 2\sqrt{2}-1\)

Dấu bằng xảy ra \(\Leftrightarrow {{\left( 1-t \right)}^{2}}=2\Leftrightarrow t=1-\sqrt{2}\,\,\left( t<0 \right)\)

\(\Rightarrow \sin x+\cos x=1-\sqrt{2}\Leftrightarrow \sqrt{2}\sin \left( x+\frac{\pi }{4} \right)=1-\sqrt{2}\Leftrightarrow \sin \left( x+\frac{\pi }{4} \right)=\frac{1-\sqrt{2}}{2}\)

Chọn A.