Question

Solve :
If $\sqrt{\sin x}+\cos x=0$, then the value of $\sin x$?

Answer 1

Given,

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

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

squaring on both sides,

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

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

for converting into a perfect square, we add $\frac{1}{4}$,

${\sin }^{2}x+\sin x+\frac{1}{4}=1+\frac{1}{4}$

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

$\therefore \sin x+\frac{1}{2}=\pm \frac{\sqrt{5}}{2}$

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

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

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