3A. Solve the equation

$$
\sin ^{5} x+\cos ^{5} x+\sin ^{4} x=2
$$
Solution. We will write the equation in the form $\sin ^{5} x+\cos ^{5} x=2-\sin ^{4} x$. Note that $\cos ^{5} x \leq \cos ^{2} x, \sin ^{5} x \leq \sin ^{2} x$. Therefore, for the left side of the last equation, we have

$$
\sin ^{5} x+\cos ^{5} x \leq \sin ^{2} x+\cos ^{2} x=1
$$

On the other hand, for the right side, we have $2-\sin ^{4} x \geq 1$, so the given equation is equivalent to the system of equations:

$$
\left\{\begin{array}{l}
\sin ^{5} x+\cos ^{5} x=1 \\
2-\sin ^{4} x=1
\end{array}\right.
$$

From the second equation, we have $\sin x= \pm 1$, so

$$
\cos ^{5} x=0 \text { or } \cos ^{5} x=2
$$

From the last equations, we obtain the desired solutions in the form

$$
x=\frac{\pi}{2}+2 k \pi, \quad k \in \mathbb{Z}
$$

4AB. The side $B C$ of triangle $A B C$ has length $a$ and the radius of the inscribed circle is $r$. Determine the area of the triangle if the inscribed circle touches the circle with diameter $B C$.