Let's calculate the value of $\operatorname{tg} x$ if $\operatorname{tg} x = \cos x$.

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Let's calculate the value of $\operatorname{tg} x$ if $\operatorname{tg} x = \cos x$.
A $\operatorname{tg} x=\frac{\sin x}{\cos x}$ relationship leads us to the following equivalent equation:

$$
\sin x=\cos ^{2} x \quad\left(x \neq \frac{\pi}{2}+k \pi, k \text { any integer }\right)
$$

or using the identity $\sin ^{2} x+\cos ^{2} x=1$,

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

From this, the possible values of $\sin x$ are: $(\sin x)_{1,2}=\frac{-1 \pm \sqrt{5}}{2}$. Given that $|\sin x| \leqslant 1$, only $\sin x=\frac{\sqrt{5}-1}{2}$ is valid. Since $\cos ^{2} x=\sin x$, we have $\cos x= \pm \sqrt{\frac{\sqrt{5}-1}{2}}$. These are also the sought values of $\operatorname{tg} x$. Our result means that if $\operatorname{tg} x=\cos x$, then $\operatorname{tg} x=\sqrt{\frac{\sqrt{5}-1}{2}}$, or $\operatorname{tg} x=-\sqrt{\frac{\sqrt{5}-1}{2}}$, but it does not mean that if $\operatorname{tg} x$ equals any of these values, then $\operatorname{tg} x=\cos x$ (we can only state that in this case $\cos x= \pm \operatorname{tg} x$).