Explanation
$${\textbf{Step -1: Rewrite the given equation as quadratic equation in Sinx by applying Standard Trigonometric Identity}}$$
$${\text{Consider,}}$$
$$\sqrt {\sin x} + \cos x = 0$$
$$\Rightarrow \sqrt {\sin x} = - \cos x$$
$${\text{Squaring on both sides, we get}}$$
$${\left( {\sqrt {\sin x} } \right)^2} = {\left( { - \cos x} \right)^2}$$
$$ \Rightarrow \sin x = {\cos ^2}x \ldots \left( 1 \right)$$
$${\text{As we know that,}}$$
$${\sin ^2}x + {\cos ^2}x = 1$$
$$ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$$ $${\textbf{(Substitute this value in equation }}\left( 1 \right))$$
$$ \Rightarrow \sin x = 1 - {\sin ^2}x$$
$$ \Rightarrow {\sin ^2}x + \sin x - 1 = 0 \ldots \left( 2 \right)$$
$${\textbf{Step -2: Solving equation}}\left( \mathbf 2 \right)$$ $${\textbf{for}}$$ $$\textbf {sin} \mathbf{x.}$$
$${\text{If }}$$ $$a{x^2} + bx + c = 0$$ $${\text{is a quadratic equation then roots of the equation are,}}$$
$$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$
$${\text{So, comparing it with equation }}\left( 2 \right)$$ $${\text{we get,}}$$
$$a = 1,b = 1,c = - 1$$
$$\therefore \sin x = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$$
$$ \Rightarrow \sin x = \dfrac{{ - 1 \pm \sqrt 5 }}{2}$$
$$ \Rightarrow \sin x = \dfrac{{ - 1 - \sqrt 5 }}{2}$$ $${\text{or}}$$ $$\sin x = \dfrac{{ - 1 + \sqrt 5 }}{2}$$
$$ \Rightarrow \sin x = \dfrac{{\sqrt 5 - 1}}{2}$$
$$\because \dfrac{{ - 1 - \sqrt 5 }}{2} < - 1,{\text{Therefore, the other value is unacceptable}}{\text{.}}$$
$${\textbf{Hence, option (D)}}$$ $$\mathbf{\dfrac{{\sqrt 5 - 1}}{2}}$$ $${\textbf{is the correct answer.}}$$
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