First try to find the solutions over $\mathbf R$, then select those which in $[0,2\pi]$.
Set $t=\tan \dfrac x 2 (x\not\equiv\pi\mod 2\pi)$. The restriction on the values of $x$ is unimportant, as one checks these values are no solutions of the equation. We know that
$$\cos x=\frac{1-t^2}{1+t^2},\quad\sin x=\frac{2t}{1+t^2}.$$
Replacing in $\,\cos x(1-2\sin x)=1$ leads us, after some simplifications, to:
$$2t^2(t-1)^2=0\iff \begin{cases}t=0\\t=1\end{cases}$$
Now
- $t=\tan\dfrac x2=0\iff \dfrac x2\equiv 0\mod\pi\iff x\equiv 0\mod 2\pi$,
- $t=\tan\dfrac x2=1\iff \dfrac x2\equiv \dfrac\pi 4\mod\pi\iff x\equiv \dfrac\pi 2\mod 2\pi$.
In the prescribed interval, we find only three solutions:
$$x\in\Bigl\{0,\,2\pi,\,\frac\pi 2\Bigr\}$$
Note that equivalently $\sin 3\theta = -\frac{1}{2}$ is also satisfied for $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$, and full revolutions from these points. We can therefore have that
$$
3\theta \;\; =\;\; \frac{7\pi}{6} + 2\pi k \hspace{2pc} \text{or} \hspace{2pc} 3\theta \;\; =\;\; \frac{11\pi}{6} + 2\pi k
$$
where $k \in \mathbb{Z}$. These equations can be rewritten as
$$
\theta \;\; =\;\; \frac{(7 + 12k)\pi}{18} \hspace{2pc} \text{or} \hspace{2pc} \theta \;\; =\;\; \frac{(11+12k)\pi}{18}.
$$
To find which values are contained in $[0,2\pi)$, we see which values of $k$ allow the above quantities to be $\geq 0$ but $<2\pi$. The solution set is found to be
$$
\left \{\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{19\pi}{18}, \frac{23\pi}{18}, \frac{31\pi}{18}, \frac{35\pi}{18} \right \}.
$$
These values are obtained by plugging in $k=0,1,2$ into the above equations. You can verify that these are all less than $2\pi$ and the only solutions which are nonnegative.
Best Answer
Arturo's hint already tells you how to proceed with the computations. To check whether the solutions that you calculate seem plausible, and to make sure that you haven't missed any solutions, it's useful to write the equation as $$\cos 2x = \sin x + 1.$$ Then you can easily plot the curves $y=\cos 2x$ (which looks like the familiar $y=\cos x$ except that it oscillates twice as fast) and $y=\sin x+1$ (the familiar sine curve, but moved one unit upwards), and see roughly where they intersect (which they do at the $x$-values that satisfy the equation).
(It should look like this, but try it yourself before you peek!)