Verify that the given function $y = \cos x + C$ is a solution of the differential equation $y^{\prime} + \sin x = 0$.

(N/A) Given function: $y = \cos x + C$
Differentiating both sides with respect to $x$,we get:
$y^{\prime} = \frac{d}{dx}(\cos x + C)$
$y^{\prime} = -\sin x$
Now,substitute the value of $y^{\prime}$ into the given differential equation $y^{\prime} + \sin x = 0$:
$L.H.S. = y^{\prime} + \sin x$
$L.H.S. = -\sin x + \sin x$
$L.H.S. = 0$
Since $L.H.S. = R.H.S.$,the given function $y = \cos x + C$ is indeed a solution of the differential equation $y^{\prime} + \sin x = 0$.