Question

If $\sqrt{\sin x}+\cos x=0$, then $\sin x=$

• A. $\frac{\sqrt{5}+1}{2}$
• B. $\frac{\sqrt{5}+1}{8}$
• C. $\frac{\sqrt{5}-1}{8}$
• D. $\frac{\sqrt{5}-1}{2}$

Answer 1

The correct option is D $\frac{\sqrt{5}-1}{2}$

$\sqrt{\sin }x+\cos x=0$

$\sqrt{\sin x}=-\cos x$

Squaring both sides:
$\sin x={\cos }^{2}x$

$=1-{\sin }^{2}x$

${\sin }^{2}x+\sin x-1=0$

Using quadratic formula to find the roots of a quadratic equation, we get:
$\sin x=\frac{-1\pm \sqrt{1+4}}{2}$

$\sin x=\frac{\sqrt{5}-1}{2}$ (The other value is unacceptable as it is less than -1.)