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:
$x= \frac{7\pi}{6} + 2\pi k,\frac{11\pi}{6} + 2\pi k, \frac{\pi}{2} + 2\pi k$

$2\cos^2(x)+\sin(x)=1$
Using $2\cos^2(x)-1=\cos(2x)$ we have $\cos(2x)=-\sin(x)=\cos(\pi/2+x).$
Hence $2x=\pm(\pi/2+x)+2n\pi$, $n\in\mathbb{Z}$ and $x=\pi/2+2n\pi$ or $x=-\pi/6+2n\pi/3.$ The second expression can be re-written as $-\pi/6\pm2\pi/3+2k\pi$ or $-\pi/6+2k\pi$ giving the three solutions 
\begin{align}
x&=\pi/2+2n\pi\\x&=-\pi/6+2k\pi=11\pi/6+2k\pi\\x&=-5\pi/6+2k\pi=7\pi/6+2k\pi\\
\end{align}
Note that one of the solutions from the second expression, $-\pi/6+2\pi/3+2k\pi=\pi/2+2k\pi$ is absorbed into the first of the three solutions. This happens because this is a double root, tangential to the x-axis - see plot from Wolfrom Alpha:

