Why squaring the trigonometric equation changes the solution?

I have a trigonometric equation that is defined as:

$\sin(\alpha) - \cos(\alpha) = \frac{1}{2}$

Solving this equation by mathematica will yield $\alpha = 65.70 $ and $\alpha = -155.705 $

But As I solve it analytically, I will obtain different results:

First I exponentiate both sides to the power of two:

$(\sin(\alpha) - \cos(\alpha))^2 = \frac{1}{2}^2$

Now I expand the expressions:

$\sin(\alpha)^2 - 2 \sin(\alpha) \cos(\alpha) + \cos(\alpha)^2 = \frac{1}{4}$

As $\sin(\alpha)^2 + \cos(\alpha)^2 = 1$, I will have:

$1 - 2 \sin(\alpha) \cos(\alpha) = \frac{1}{4}$

Again if I put $2 \sin(\alpha) \cos(\alpha) = \sin(2\alpha)$ I will be left with:

$\sin(2 \alpha) = \frac{3}{4}$

Which will readily give $\alpha = 24.3$

So, why I am having different results? I don't understand.

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).