Question

The smallest positive root of the equation $\sqrt{sin(1-x)}=\sqrt{cosx}$, is-

• A. $\frac{1}{2}+\frac{7\pi }{4}$
• B. $\frac{1}{2}+\frac{5\pi }{4}$
• C. $\frac{1}{2}+\frac{3\pi }{4}$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{2}+\frac{7\pi }{4}$

To find the smallest positive root of the equation is
$\sqrt{sin(1-x)}=\sqrt{cosx}$
On Squaring both side , we get
$sin(1-x)=cosx$
$sin(1-x)=sin(x-\frac{7\Pi }{2})$
$1-x=x-\frac{7\Pi }{2}$
$2x=1+\frac{7\Pi }{2}$
$x=\frac{1}{2}+\frac{7\Pi }{4}$
Hence , Option A