B3. Determine the domain of the function $f(x)=\ln \frac{1}{\cos x}$ and write down the points where the function $f(x)$ has tangents parallel to the angle bisectors of the even quadrants.
B3. For $f(x)=\ln \frac{1}{\cos x}$ to be defined, it must hold that $\frac{1}{\cos x}>0$ or $\cos x>0$. The domain is thus the union of intervals of the form $\left(-\frac{\pi}{2}+2 k \pi, \frac{\pi}{2}+2 k \pi\right)$ for all integers $k$.

The tangent will be parallel to the bisectors of the even quadrants if its slope is -1, i.e., $f^{\prime}(x)=-1$. It follows that $-1=\frac{1}{\frac{1}{\cos x}} \cdot \frac{-1}{\cos ^{2} x}(-\sin x)=\operatorname{tg} x$, so the valid $x$ are those from the domain for which $x \stackrel{\pi}{=}-\frac{\pi}{4}+2 k \pi$. For these $x$, it also holds that $f(x)=\ln \left(\frac{1}{\cos \left(\frac{\pi}{4}\right)}\right)=\frac{1}{2} \ln 2$. We get the points $T\left(-\frac{\pi}{4}+2 k \pi, \frac{1}{2} \ln 2\right)$.