How to choose between two solutions where both seem to be correct? Question to solve: $ y\sin (2x)dx - (y^2 + \cos ^2x)dy = 0 $ -----(i) 
Writing (i) in exact form [$ M(x,y)dx + N(x,y)dy = 0$], we get
$y\sin (2x)dx + [-(y^2 + \cos ^2x)]dy = 0$ $where,M=y\sin (2x)$ and $N=[-(y^2 + \cos ^2x)]$

Since the given differential equation is exact, i.e, $∂M/∂y = ∂N/∂x = \sin (2x)$
it's general solution is given by 
$∫Mdx$ (with y constant) + $∫Ndy $ ("N" free from x terms) $=C$
My workout:
$∫Mdx$ (with y constant) = $∫y\sin (2x)dx$ = $\frac{-y\cos (2x)}2$
$∫Ndy $ ("N" free from x terms)$ = ∫-y^2 dy = \frac{-y^3}3 $
So, the general solution is $\frac{-y\cos (2x)}2$ + $\frac{-y^3}3 $ $= C$
But if we differentiate the general solution it does not trace back to original Differential Equation.
However if I simplify $ \cos ^2(x)$ as $\frac{\cos 2x+1}2$ I receive [$\frac{-y\cos (2x)}2$ + $\frac{-y^3}3 - \frac{y}2$ $= C$] as general solution and it does trace back to original differential eqaution. So, the problem is do i need verify everytime whether the obtained solution is correct or not, or both of the solutions are correct here?

The following line in your original solution is wrong, which is the reason the answer you obtained is wrong (and thus doesn't satisfy the original equation):

$\int Ndy$ ("N" free from x terms)$= ∫-y^2 dy  = \frac{-y^3}3$

A few lines earlier you stated (correctly!) that $N=[-(y^2+\cos^2x)]$, therefore the integral of $N$ is
$$\int N\,dy=\int[-(y^2+\cos^2x)]\,dy,$$
not what you set up above, as the function you put inside the integral is not $N$.
