Solve the following equation:

$$
\sqrt{2} \cdot(\sin x+\cos x)=\tan x+\cot x
$$
Solution. The left side of the equation, according to the sum formula of the cosine function, is

$$
\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x=\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cos x=\cos \left(\frac{\pi}{4}-x\right)
$$

therefore, the absolute value of the left side of the equation is at most 2. If the right side is meaningful, then $\operatorname{ctg} x=1 / \operatorname{tg} x$ and as the sum of a real number and its reciprocal, the absolute value is at least 2. Thus, for every solution $x$ of the equation, $\left|\cos \left(\frac{\pi}{4}-x\right)\right|=1$, which means $x=\frac{\pi}{4}+k \pi$ holds for some integer $k$. Transforming the right side identically:

$$
\operatorname{tg} x+\operatorname{ctg} x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}=\frac{2}{\sin 2 x}
$$

which is 2 when $x=\frac{\pi}{4}+l \pi($ where $l \in \mathbb{Z})$. Therefore, $\cos \left(\frac{\pi}{4}-x\right)=1$, that is, only the angles of the form

$$
x=\frac{\pi}{4}+2 l \pi
$$

are solutions to the equation.