11.6. Solve the equation

$$
\cos 2x - \sin 2x + 2 \cos x + 1 = 0
$$
Solution. We have the equivalent equations:

$$
\begin{gathered}
\cos 2 x-\sin 2 x+2 \cos x+1=0 \\
1-2 \sin ^{2} x-2 \sin x \cos x+2 \cos x+1=0 \\
\sin ^{2} x+\sin x \cos x-\cos x-1=0 \\
\left(\sin ^{2} x-\sin x\right)+(\sin x \cos x-\cos x)+(\sin x-1)=0 \\
(\sin x-1) \cdot(\sin x+\cos x+1)=0 \Leftrightarrow \\
{\left[\begin{array}{c}
\sin x-1=0 \\
\sin x+\cos x+1=0
\end{array}\right.}
\end{gathered}
$$

For the second equation, we have the following equivalent equations:

$$
\begin{gathered}
\sin x+\cos x+1=0 \\
2 \sin \frac{x}{2} \cos \frac{x}{2}+2 \cos ^{2} \frac{x}{2}=0 \\
\cos \frac{x}{2}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)=0 \Leftrightarrow \\
{\left[\begin{array} { c } 
{ \operatorname { cos } \frac { x } { 2 } = 0 } \\
{ \operatorname { sin } \frac { x } { 2 } + \operatorname { cos } \frac { x } { 2 } = 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\cos \frac{x}{2}=0 \\
\operatorname{tan} \frac{x}{2}=-1
\end{array}\right.\right.}
\end{gathered}
$$

Thus, the initial equation has been reduced to three equations: 1) $\sin x=1$, 2) $\cos \frac{x}{2}=0$ or 3) $\operatorname{tan} \frac{x}{2}=-1$.

The equation $\sin x=1$ has solutions $x=\frac{\pi}{2}+2 k \pi, k \in \mathbf{Z}$.

The equation $\cos \frac{x}{2}=0$ has solutions $x=\pi+2 m \pi, m \in \mathbf{Z}$.

The equation $\operatorname{tan} \frac{x}{2}=-1$ has solutions $x=-\frac{\pi}{2}+2 n \pi, n \in \mathbf{Z}$.

Therefore, the solutions to the initial equation are $x=x_{0}+2 k \pi, k \in \mathbf{Z}, x_{0} \in\left\{-\frac{\pi}{2}, \frac{\pi}{2}, \pi\right\}$.