Put $=\sin x+\cos x=p \Rightarrow 1+2 \sin x \cos x=p^{2}$

$\sin x \cos x=\left(\frac{p^{2}-1}{2}\right)$

$\therefore f(x) = \left| {p + \frac{2}{{\left( {{p^2} – 1} \right)}} + \frac{{2p}}{{{p^2} – 1}}} \right|$

${=\left|p+\frac{2}{(p-1)}\right|} $

${=\left|(p-1)+\frac{2}{(p-1)}+1\right|}$

$\therefore \frac{(p-1)+\left(\frac{2}{p-1}\right)}{2} \geq\left((p-1) \cdot \frac{2}{p-1}\right)^{1 / 2}$

$\therefore(p-1)+\left(\frac{2}{p-1}\right) \geq 2 \sqrt{2}$

$\therefore \quad f(x)_{\min } \geq 2 \sqrt{2}+1$