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:
$x= \frac{7\pi}{6} + 2\pi k,\frac{11\pi}{6} + 2\pi k, \frac{\pi}{2} + 2\pi k$
 A: 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 $t=\sin x$
$$2-2 \sin^2 x+\sin x=1 \iff 2t^2-t-1=0$$
$$t_{1,2}=\frac{1\pm \sqrt{9}}{4}=1, -\frac12$$
A: Your method's fine, the answer's right. The only improvement I can suggest is to make the definition of "base solution" clear upfront. Each "and so" acts as if a specific value of $\sin x$ has finitely many solutions rather than finitely many per period, so before you obtain them you should mention a restriction to $[0,\,2\pi)$ and then extend to $\Bbb R$ at the end. 
A: Other way is used identities double angle and sum-product
\begin{eqnarray*} 
2\cos^2x+\sin x& = & 1 \\ 
2\cos^2x-1+\sin x& = & 0\\
\cos(2x)+\sin x& = & 0\\
\cos(2x)+\cos\left(\frac{\pi}{2}-x\right) & = & 0\\
2\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)\cos\left(\frac{3x}{2}-\frac{\pi}{4}\right) & = & 0\\
  \end{eqnarray*} 
A: $2\cos^2(x)+\sin(x)=1$
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:

