The smallest positive root of the equation 

$\sqrt{\sin (1-x)}=\sqrt{\cos x}$ is 

$\sin (1-x)\ge 0\cdot \cos x\ge 0$ and $\sin (1-x)=\cos x$ 

$\Rightarrow 1-x=n\pi +(-1{)}^n(\pi /2-x)$ where $n\in I$

For $n=2m$ we get no values of $x,m\in I$ 

$\therefore 1-x=(2m+1)\pi -\pi /2+x$ 

$\Rightarrow x=\frac{1}{2}-\frac{4m+1}{4}\pi$ 

for $m\ge 0,x<0$ 

For $m=-1,x=\frac{1}{2}+\frac{3\pi }{4}$

and $\sin (1-x)=\sin \left(\frac{1}{2}-\frac{3\pi }{4}\right)=-\sin \left(\frac{1}{2}+\frac{\pi }{4}\right)<0$ 

so, $x\ne \frac{1}{2}+\frac{3\pi }{4}$ 

For $m=-2,x=\frac{1}{2}+\frac{7\pi }{4}$ and 

$\sin (1-x)=\sin \left(\frac{1}{2}+\frac{\pi }{4}-2\pi \right)>0$ and 

$\cos x=\cos \left(\frac{1}{2}+2\pi -\frac{\pi }{4}\right)>0$

Hence $x=\frac{1}{2}+\frac{7\pi }{4}$ is the smallest positive root of 

the equation.