Question

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

• A. ${\sin }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
• B. $\pi -{\sin }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
• C. ${\sin }^{-1}\left(\frac{1-\sqrt{5}}{2}\right)$
• D. ${\cos }^{-1}\left(\sqrt{\frac{\sqrt{5}-1}{2}}\right)$

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$

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

$\cos x=\sqrt{sinx}$

Squaring on both sides

${\cos }^{2}x=\sin x$

$[\therefore {\cos }^{2}x=1-{\sin }^{2}x]$

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

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

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

Let $\sin x=y$

$y^2+y-1$

$\frac{-1\pm \sqrt{(1{)}^2-4(1\left)\right(-1)}}{2\left(1\right)}$

$y=\frac{-1\pm \sqrt{5}}{2}$

$\sin x\ne \frac{-1-\sqrt{5}}{2}$ $|$ $\sin x=\frac{-1+\sqrt{5}}{2}$

$\ne -1.6$

since $\sin x$ lies between $-1$ and $1$ $\therefore \sin x=\frac{\sqrt{5}-1}{2}$

$x={\sin }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$.