Question

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

• A. $x=-\frac{1}{2}+\frac{7\pi }{4}$, is the smallest positive root of the given equation
• B. $x=\frac{1}{2}+\frac{7\pi }{4}$, is the smallest positive root of the given equation
• C. $x=\frac{1}{2}+\frac{5\pi }{4}$, is the smallest positive root of the given equation
• D. $x=-\frac{1}{2}+\frac{3\pi }{4}$, is the smallest positive root of the given equation

Answer 1

The correct option is B $x=\frac{1}{2}+\frac{7\pi }{4}$, is the smallest positive root of the given equation

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