3.375. Find $\operatorname{tg} \frac{x}{2}$, if it is known that $\sin x+\cos x=\frac{1}{5}$.
Solution.

Let $\sin x + \cos x = \frac{1}{5}$.

Since $\sin \alpha = \frac{2 \operatorname{tg} \frac{\alpha}{2}}{1 + \operatorname{tg}^{2} \frac{\alpha}{2}} ; \cos \alpha = \frac{1 - \operatorname{tg}^{2} \frac{\alpha}{2}}{1 + \operatorname{tg}^{2} \frac{\alpha}{2}}, \alpha \neq (2 n + 1) \pi, n \in Z$, then

we have $\frac{2 \operatorname{tg} \frac{x}{2}}{1 + \operatorname{tg}^{2} \frac{x}{2}} + \frac{1 - \operatorname{tg}^{2} \frac{x}{2}}{1 + \operatorname{tg}^{2} \frac{x}{2}} = \frac{1}{5} \Leftrightarrow 3 \operatorname{tg}^{2} \frac{x}{2} - 5 \operatorname{tg} \frac{x}{2} - 2 = 0$, from which, solving this
equation as a quadratic equation in $\operatorname{tg} \frac{x}{2}$, we find $\operatorname{tg} \frac{x}{2} = -\frac{1}{3}$ or

$\operatorname{tg} \frac{x}{2} = 2$

Answer: $\operatorname{tg} \frac{x}{2} = 2$ or $\operatorname{tg} \frac{x}{2} = -\frac{1}{3}$.