$\pi\leq x \leq 2\pi$

$\Rightarrow 180^o\leq x \leq 360^o$

Vậy x thuộc góc phần tư thứ III hoặc IV

$C=\sqrt{2}-\frac{1}{sin(x+2013\pi)}.\sqrt{\frac{1}{1+cosx}+\frac{1}{1-cosx}}$

 $=\sqrt{2}-\frac{1}{sin(x+\pi+2012\pi)}.\sqrt{\frac{1-cosx}{(1+cosx)(1-cosx)+\frac{1+cosx}{(1+cosx)(1-cosx)}}$

$=\sqrt{2}-\frac{1}{sin(x+\pi)}.\sqrt{\frac{2}{1+cosx-cosx-cos^x}}$

$=\sqrt{2}+\frac{1}{sinx}.\sqrt{\frac{2}{1-cos^2x}}$

$C^2=(\sqrt{2}+\frac{1}{sinx}.\sqrt{\frac{2}{1-cos^x}})^2$

$C^2=2+2\sqrt{2}\frac{1}{\sqrt{1-cos^2x}}\frac{\sqrt{2}}{\sqrt{1-cos^2x}}+\frac{1}{sin^2x}\frac{2}{1-cos^2x}$

$C^2=2+\frac{4}{1-cos^2x}+\frac{2}{(1-cos^2)^2}$

$C^2=\frac{2(1-cos^2x)^2+4(1-cos^2)+2}{(1-cos^2x)^2}$

$C^2=\frac{2(1-cos^2x)^2+4(1-cos^2)+2}{(1-cos^2x)^2}$

$C=\frac{\sqrt{2[(1-cos^2x)^2+2(1-cos^2x)+1]}}{1-cos^2x}$