If $\mathrm{\alpha }$ be the similar positive root of the equation $\sqrt{\sin (1-\mathrm{x})}=\sqrt{\mathrm{cosx}}$, then the approximate integral value of $\mathrm{\alpha }$ is equal to

The given equation is possible if $\sin (1-\mathrm{x})\ge 0$ and $\mathrm{cosx}\ge 0$. From the given equation, on squaring both the sides we get $\sin (1-\mathrm{x})=\mathrm{cosx}=\sin (\frac{\mathrm{\pi }}{2}-\mathrm{x})$
$\Rightarrow 1-\mathrm{x}=\mathrm{n\pi }+{(-1)}^{\mathrm{n}}(\frac{\mathrm{\pi }}{2}-\mathrm{x})$ where $(\mathrm{n}\in \mathrm{I})$ But for $\mathrm{n}=2\mathrm{m},(\mathrm{n}\in \mathrm{I})$ we get no value of
$\therefore 1-\mathrm{x}=(2\mathrm{m}+1)\mathrm{\pi }-\frac{\mathrm{\pi }}{2}+\mathrm{x}\Rightarrow \mathrm{x}=\frac{1}{2}-\frac{4\mathrm{m}+1}{4}\mathrm{\pi }$ Now, $\mathrm{m}\ge 0\Rightarrow \mathrm{x}<0$
if $\mathrm{m}=-1\Rightarrow \mathrm{x}=\frac{1}{2}+\frac{3\mathrm{\pi }}{4}\Rightarrow \sin (1-\mathrm{x})=\sin \left(\frac{1}{2}-\frac{3\mathrm{\pi }}{4}\right)=-\sin \left(\frac{1}{2}+\frac{\mathrm{\pi }}{4}\right)<0$
Hence rejectedIf $\mathrm{m}=-2\Rightarrow \mathrm{x}=\frac{1}{2}+\frac{7\mathrm{\pi }}{4}$
$\Rightarrow \sin (1-\mathrm{x})=\sin (\frac{1}{2}+\frac{\mathrm{\pi }}{4}-2\mathrm{\pi })>0$ and $\mathrm{cosx}=\cos \{\frac{1}{2}-(\frac{\mathrm{\pi }}{4},\frac{1}{2})\}$
Hence, $\mathrm{x}=\frac{1}{2}+\frac{7\mathrm{\pi }}{4}$ is the smallest +ve root of the given equation.
So, $\mathrm{\alpha }=\frac{1}{2}+\frac{7\mathrm{\pi }}{4}=\frac{1}{2}+\frac{7}{4}\times \frac{22}{7}=6$