You actually nailed it, you just didn’t know it.
Algebra’s fundamental theorem states: Any $\rm n^{th}$ order polynomial has $\rm n$ roots.
Then, if one increase the order of a polynomial by $\rm m$ (let’s say $\rm 1^{st}$ to $\rm 2^{nd}$) the number of roots will increase from $\rm n$ to $\rm n+m$, Therefore raising the power makes things different.
But raising to the same power both sides of an equation makes the solution set of the $\rm (n+m)^{th}$ order polynomial to contain the solutions set of the $\rm n^{th}$ polynomial. Meaning, it shouldn’t be a problem in your case, your solutions set should be in the square problem.
Start to get the solutions of $\sin(2α)=\frac{1}{2}$. If $0 ≤α ≤360º$, then your solution set will be $\{24.3º, 65.70º, 204.295º (-155.705º), 245.365º\}$ (check the graph below).

Values in the middle are the ones you are looking for (see graph below).

Has you can see by this graph, the actual set of solutions is much bigger. Let’s find the full solution set to your problem. To do so, select the solutions of the square problem which are also a solution to your initial problem in the range $\rm 0≤α≤360$ i.e., $\rm α_1=65.7º, α_2=204.295º$. Now, let’s see how many times the $\sin$ of $2×65.7º$ or $\sin (131.4º)$ repeats.

If we imagine a circle it becomes clear that after every full turn one gets back to same ($\rm 65.7º$) point, i.e., the first set of solutions satisfy
$$\rm \Bbb α_{1n}=2nπ+α_1 \approx n×360º+65.7º, \forall n \in \Bbb N$$
But, we can check that every $180º-65.7º$ has the same $\sin$ of $65.7º$. Appling previous reasoning, the second set satisfy
$$\rm \Bbb α’_{1n}=(2n+1)π-α_1 \approx n×360º+180º-65.7º=n×360º+114.3º, \forall n \in \Bbb N$$
We do the same with $\rm α_2$. The result is
$$\rm \Bbb α_{2n}=2nπ+α_2 \approx n×360º+204.295º, \forall n \in \Bbb N$$
$$\rm \Bbb α’_{2n}=(2n+1)π-α_2 \approx n×360º+180º-204.295º=n×360º-24,295º, \forall n \in \Bbb N$$
The total set of solutions is:
$$\rm \Bbb S= α_{1n} \cup α’_{1n} \cup α_{2n} \cup α’_{2n}, \forall n \in \Bbb N$$