Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$ $2\cos^2 x+\sin x=1$
$\Rightarrow 2(1-\sin^2 x)+\sin x=1$
$\Rightarrow 2-2 \sin^2 x+\sin x=1$
$\Rightarrow 0=2 \sin^2 x- \sin x-1$
And so:
$0 = (2 \sin x+1)(\sin x-1)$
So we have to find the solutions of each of these factors separately:
$2 \sin x+1=0$
$\Rightarrow \sin x=\frac{-1}{2}$
and so $x=\frac{7\pi}{6},\frac{11\pi}{6}$
Solving for the other factor,
$\sin x-1=0 \Rightarrow \sin x=1$
And so $x=\frac{\pi}{2}$

Now we have found all our base solutions, and so ALL the solutions can be written as so:
$x= \frac{7\pi}{6} + 2\pi k,\frac{11\pi}{6} + 2\pi k, \frac{\pi}{2} + 2\pi k$

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 $t=\sin x$
$$2-2 \sin^2 x+\sin x=1 \iff 2t^2-t-1=0$$
$$t_{1,2}=\frac{1\pm \sqrt{9}}{4}=1, -\frac12$$
