Solve the following equation:

$\sin x + \cos x + \sin x \cos x = 1$.
I. Solution. It is clear that both trigonometric functions cannot be negative. We will show that neither can be. If, for example, $\sin x < 0, \cos x \geq 0$, then the left side can be written as

$$
\sin x + \cos x(1 + \sin x)
$$

with $0 \leq 1 + \sin x < 1$, so the absolute value of each term is less than 1, and the first term is negative, making the expression less than 1. Similarly, by swapping the roles of $\sin x$ and $\cos x$, it can be shown that $\cos x$ cannot be negative either.

If $\sin x \geq 0, \cos x \geq 0$, then - since a number between 0 and 1 is not less than its square -

$$
\sin x + \cos x + \sin x \cos x \geq \sin x + \cos x \geq \sin^2 x + \cos^2 x = 1
$$

Here, equality must hold everywhere for (1) to be satisfied. In the first case, equality holds if either $\sin x = 0$ - and in this case, due to $\cos x \geq 0$, $\cos x = 1$ - or $\cos x = 0, \sin x = 1$. In these cases, (1) is indeed satisfied, so we have found all solutions to the equation. The corresponding angle values are:

$$
x = k \cdot 360^{\circ}, (k = 0, \pm 1, \pm 2, \ldots) \text{ and } x = 90^{\circ} + k \cdot 360^{\circ} (k = 0, \pm 1, \pm 2, \ldots).
$$

II. Solution. Move the third term on the left side to the right side and square both sides:

$$
\sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 - 2 \sin x \cos x + \sin^2 x \cos^2 x
$$

The sum of the first two terms on the left side is 1, so the equation can be rearranged as:

$$
\sin x \cos x (4 - \sin x \cos x) = 0
$$

The value of the last factor is at least 3, so the equation can only be satisfied if $\sin x = 0$ or $\cos x = 0$. Substituting these into (1) gives that $\cos x = 1$ or $\sin x = 1$ must be true. Such angles $x$ indeed exist:

$$
\begin{array}{ll}
x = k \cdot 360^{\circ} & (k = 0, \pm 1, \pm 2, \ldots), \text{ or } \\
x = 90^{\circ} + k \cdot 360^{\circ} & (k = 0, \pm 1, \pm 2, \ldots).
\end{array}
$$