Why squaring the trigonometric equation changes the solution?
If you square, you also get the solutions of $$ \sin\alpha-\cos\alpha=-\frac{1}{2} $$
In general, if you have an equation of the form $f(x)=g(x)$ and square both sides, you get, after rearranging, $$ f(x)^2-g(x)^2=0 \tag{*} $$ that can be rewritten $$ (f(x)-g(x))(f(x)+g(x))=0 $$ so the solutions of (*) are the solutions of $f(x)-g(x)=0$ (the original equation) together with the solutions of $f(x)+g(x)=0$.
A safer way to solve your equation is to set $t=\tan(\alpha/2)$, so the equation becomes $$ \frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}=\frac{1}{2} $$ that reduces to $$ t^2+4t-3=0 $$ with solutions $$ -2+\sqrt{7}\qquad\text{or}\qquad -2-\sqrt{7} $$ The first solutions corresponds to $$ \alpha=2\arctan(\sqrt{7}-2)\approx 65.7^\circ $$ and the second one to $$ \alpha=-2\arctan(\sqrt{7}+2)\approx-155.7^\circ $$ or $\approx204.3^\circ$ if you want a value between $0$ and $360$.
Notice first of all that besides $24.3°$ you also have the solution $90°-24.3°=65.7°$ (supplementary angles have the same sine).
Your solution $24.3°$ is to discard, because it is a solution of the equation $\cos\alpha-\sin\alpha=1$ (which of course is the same as the given one when squared).
In summary: when you solve $\sin2\alpha=3/4$ you must consider all solutions, and discard fake ones:
$\alpha=48.6°/2$ (discard)
$\alpha=(48.6°+360°)/2$ (fine, it is the same as $-155.7°$)
$\alpha=(180°-48.6°)/2$ (fine)
$\alpha=(180°-48.6°+360°)/2$ (discard).
Squaring is often a useful (sometimes unavoidable) way to solve an equation, but it always introduces new solutions which may not fit the original problem. When you square both sides of an equation, you should always check your solutions to make sure they are applicable.
If you can avoid squaring, you might try doing so so as not to be confused with extra solutions. For this particular problem, you can solve it without squaring as follows:
$\dfrac12\ =\ \sin\alpha-\cos\alpha\ =\ \sqrt2\,(\cos45^\circ\sin\alpha - \sin45^\circ\cos\alpha)\ =\ \sqrt2\sin(\alpha-45^\circ)$
which gives $\sin(\alpha-45^\circ)=\frac1{2\sqrt2}$ which you can solve to give you the values obtained from Mathematica.