Upon solving this equation:
$$ 5\sin x - 5\cos x = 2 $$
for the interval $0 \le x < 360^\circ$, I manipulated it in order to get a solvable trigonometric equation below:
$$ \sin 2x = 21/25 $$
Normally this would procure 4 solutions: $28.6^\circ, 61.4^\circ, 208.6^\circ, 241.4^\circ$ (one can easily plot this into Desmos to see that it is true).
However, the answers only include the middle 2 solutions: $61.43^\circ, 208.57^\circ$. Why is this? Why do 2 of the solutions simply disappear? Could it have something to do with the way in which I manipulated the original equation (by squaring and substituting using the appropriate trigonometric identities)?
After looking at the original equation, it makes sense to some degree, but then I'm not a fan of such methods, especially since one could encounter more complicated equations in the future.
Am I missing/forgetting something fundamental?
$$\sin{x}-\cos{x}=\frac{2}{5}$$ $$(\sin{x}-\cos{x})^2=\frac{4}{25}$$ $$1-2\sin{x}\cos{x}=\frac{4}{25}$$ $$\sin{2x}=\frac{21}{25}$$
In the second step, when we square both sides, we inadvertently include the solutions of: $$\sin{x}-\cos{x}=-\frac{2}{5}$$ This equation is true for $28.6^{\circ}$ and $241.4^{\circ}$, which are the extra two solutions which we don't need.