4.85 Find the value of $\operatorname{tg} \frac{x}{2}$, if $\sin x-\cos x=1.4$.
4.85 It is convenient to use formulas (4.28) and (4.29), considering that they are valid only for $x \neq \pi(2 n+1), n \in Z$. However, in this case, $x$ cannot take these values. Indeed, if $x=\pi(2 n+1)$, then

$$
\sin (\pi(2 n+1))-\cos (\pi(2 n+1))=0-(-1) \neq 1.4
$$

Expressing $\sin x$ and $\cos x$ in terms of $\operatorname{tg} \frac{x}{2}$, we rewrite the given equality as
$\frac{2 \operatorname{tg} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}} \div \frac{1-\operatorname{tg}^{2} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}}=1.4$.

Letting $\operatorname{tg} \frac{x}{2}=z$, we obtain the equation $z^{2}-5 z+6=0$, from which $z_{1}=2, z_{2}=3$. Thus, $\operatorname{tg} \frac{x}{2}=2$ and $\operatorname{tg} \frac{x}{2}=3$.