The following equation needs to be solved:

$$
\sin x + \cos x = \frac{\cos 2x}{1 - 2 \sin x}
$$
I. Solution: The values of $x$ that come into consideration are those for which

$$
1-\sin 2 x \neq 0, \quad \text { that is } \quad x \neq \frac{\pi}{4} \pm k \pi(k=0,1,2, \ldots)
$$

Rearranging the equation,

$$
(\sin x+\cos x)(1-2 \sin x \cos x)-\cos ^{2} x+\sin ^{2} x=0
$$

or

$$
\sin x+\cos x-2 \sin ^{2} x \cos x-2 \sin x \cos ^{2} x-\cos ^{2} x+\sin ^{2} x=0
$$

Substituting $1-\cos ^{2} x$ for $\sin ^{2} x$ everywhere and rearranging,

$$
\sin x-\cos x+2 \cos ^{3} x-2 \sin x \cos ^{2} x-2 \cos ^{2} x+1=0
$$

Factoring out $\sin x$ and $\cos x$ from the odd-degree terms,

$$
\sin x\left(1-2 \cos ^{2} x\right)+\cos x\left(2 \cos ^{2} x-1\right)-2 \cos ^{2} x+1=0
$$

The left side can be factored by $2 \cos ^{2} x-1$:

$$
\left(2 \cos ^{2} x-1\right)(\cos x-\sin x-1)=0
$$

From this, either

$$
2 \cos ^{2} x-1=0
$$

or

$$
\cos x-\sin x-1=0
$$

From (1),

$$
\cos x=\frac{1}{\sqrt{2}}
$$

or

$$
\cos x=-\frac{1}{\sqrt{2}}
$$

From (2),

$$
\cos x=1+\sin x
$$

or

$$
\cos ^{2} x=1+2 \sin x+\sin ^{2} x
$$

which, substituting $1-\sin ^{2} x$ for $\cos ^{2} x$ and rearranging, gives

$$
2 \sin ^{2} x+2 \sin x=2 \sin x(\sin x+1)=0
$$

Considering only the principal values for now,

From (3),

$$
\begin{aligned}
& x=\frac{\pi}{4} \\
& x=\frac{7 \pi}{4}
\end{aligned}
$$

From (4),

$$
\begin{aligned}
& x=\frac{3 \pi}{4} \\
& x=\frac{5 \pi}{4}
\end{aligned}
$$

From (5),

$$
\begin{gathered}
x=0 \\
x=\pi \\
x=\frac{3 \pi}{2}
\end{gathered}
$$

Since we did not always perform equivalent (equivalent) transformations (e.g., we squared), we need to check whether the roots we obtained satisfy our original equation.

The roots under (6) and (9) are precisely the excluded values. By substitution, we can verify that the root under (11) does not satisfy our equation, while the other 4 roots do indeed satisfy the equation.

Thus, the sought roots in principal values $(0 \leq x<2 \pi)$ in size order are:

$$
x_{1}=0=0^{\circ}, \quad x_{2}=\frac{3 \pi}{4}=135^{\circ}, \quad x_{3}=\frac{3 \pi}{2}=270^{\circ}, \quad x_{4}=\frac{7 \pi}{4}=315^{\circ}
$$

Of course, adding or subtracting multiples of $2 \pi=360^{\circ}$ to these principal values, we also obtain roots. Thus, all the roots (combining $x_{2}$ and $x_{4}$ into one formula):

$$
\begin{gathered}
x= \pm 2 k \pi= \pm k \cdot 360^{\circ}, x=\frac{3 \pi}{4} \pm k \pi=135^{\circ} \pm k \cdot 180^{\circ} \\
x=\frac{3 \pi}{2} \pm 2 k \pi=270^{\circ} \pm k \cdot 360^{\circ}, \quad \text { where } k=0,1,2, \ldots
\end{gathered}
$$

II. Solution: The $x$ values that come into consideration are those for which

$$
1-\sin 2 x \neq 0
$$

Square both sides of the equation, then

$$
(\sin x+\cos x)^{2}=\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x=1+\sin 2 x
$$

so in the equation, only the functions of $2 x$ will appear. Reducing the equation to 0 and transforming appropriately:

$$
\begin{gathered}
(1+\sin 2 x)(1-\sin 2 x)^{2}-\cos ^{2} 2 x=\left(1-\sin ^{2} 2 x\right)(1-\sin 2 x)-\cos ^{2} 2 x= \\
=\left(1-\sin ^{2} 2 x\right)(1-\sin 2 x)-\left(1-\sin ^{2} 2 x\right)=-\sin 2 x\left(1-\sin ^{2} 2 x\right)= \\
=-\sin 2 x(1+\sin 2 x)(1-\sin 2 x)=0
\end{gathered}
$$

Here, according to (1), the last factor cannot be 0, so only the solutions of the

$$
\sin 2 x=0 \quad \text { and } \quad \sin 2 x=-1
$$

equations come into consideration.

In the interval $0 \leq x<2 \pi$, the former is satisfied by the values $x=0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}$, and the latter by the values $x=\frac{3 \pi}{4}, \frac{7 \pi}{4}$. Of these, the original equation is satisfied by

$$
x=0, \frac{3 \pi}{2}, \frac{3 \pi}{4}, \frac{7 \pi}{4}
$$

values and, of course, by any $x$ value that differs from any of these by $\pm 2 k \pi$, where $k=1,2, \ldots$.

III. Solution: The transformations can be performed a bit more skillfully: if

$$
\cos x \neq \sin x, \text { that is } \operatorname{tg} x \neq 1, x \neq \frac{\pi}{4} \pm k \pi, \quad(k=0,1,2, \ldots)
$$

then

$$
\sin x+\cos x=\frac{\cos ^{2} x-\sin ^{2} x}{1-2 \sin x \cos x}=\frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos ^{2} x-2 \sin x \cos x+\sin ^{2} x}=\frac{\cos x+\sin x}{\cos x-\sin x}
$$

Rearranging the equation,

$$
(\sin x+\cos x)(\cos x-\sin x-1)=0
$$

so either

$$
\sin x+\cos x=0
$$

or

$$
\cos x-\sin x-1=0
$$

From (1), since $\cos x$ and $\sin x$ cannot be zero at the same time, and thus neither term in the equation can be 0,

$$
\operatorname{tg} x=-1, \quad x=-\frac{\pi}{4} \pm k \pi \quad(k=0,1,2, \ldots)
$$

