$y - \cos y = x:(y\sin y + \cos y + x)y' = y$

.: We have, $y - \cos y = x$

…(1)
Differentiating (1) w.r.t. $x$,

we get
$y' - ( - \sin y)y' = 1 \Rightarrow y'(1 + \sin y) = 1$

…(2)
Multiplying (2) by $y$,

we get
$y'y(1 + \sin y) = y \Rightarrow y'(y + y\sin y) = y$

$\Rightarrow y'(x + \cos y + y\sin y) = y$

$\therefore$ $y - \cos y = x$ is the solution of the given differential equation.