Question

Find the smallest positive roots of the equation $\sqrt{\sin (1-x)}=\sqrt{\cos x}$

• A. $x=\frac{7\pi }{4}-\frac{1}{2}$.
• B. $x=\frac{7\pi }{4}+\frac{1}{2}$.
• C. $x=\frac{5\pi }{4}+\frac{1}{2}$.
• D. $x=\frac{5\pi }{4}-\frac{1}{2}$.

Answer 1

The correct option is B $x=\frac{7\pi }{4}+\frac{1}{2}$.

Given $\sqrt{\sin (1-x)}=\sqrt{\cos x}$

Squaring both side by taking care that $\sin (1-x),\cos x\ge 0$

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

$\Rightarrow x=2n\pi -(\frac{\pi }{2}-1+x)$, where $nϵZ$

$\Rightarrow x=(4n-1)\frac{\pi }{4}+\frac{1}{2}$

For smallest positive root take $n=1$, but for this $\cos x<0$

Thus taking $n=2$

$\Rightarrow x=\frac{7\pi }{4}+\frac{1}{2}$