Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$

Yes your solution is very nice and correct, as a slightly different alternative

$$2\cos^2(x)+\sin (x)=1 \iff 2(1-\sin x)(1+\sin x)+\sin x-1=0 $$

$$\iff (\sin x-1)(-2-2\sin x)+(\sin x-1)=0 \iff (\sin x-1)(-1-2\sin x)=0$$

which indeed leads to the same solutions, or also from here by $t=\sin x$

$$2-2 \sin^2 x+\sin x=1 \iff 2t^2-t-1=0$$

$$t_{1,2}=\frac{1\pm \sqrt{9}}{4}=1, -\frac12$$