Solve the equation

$$
\sin ^{5} x+\cos ^{5} x+\sin ^{4} x=2
$$
Solution. Since $\sin x \leq 1, \cos x \leq 1$ and $\sin ^{2} x+\cos ^{2} x=1$ holds for any value of $x$, therefore:

$$
\begin{gathered}
\sin ^{5} x+\cos ^{5} x+\sin ^{4} x=\sin ^{3} x \cdot \sin ^{2} x+\cos ^{3} x \cdot \cos ^{2} x+\sin ^{4} x \leq \\
\leq 1^{3} \cdot \sin ^{2} x+1^{3} \cdot \cos ^{2} x+1^{4}=1 \cdot\left(\sin ^{2} x+\cos ^{2} x\right)+1=2
\end{gathered}
$$

We obtained that $\sin ^{5} x+\cos ^{5} x+\sin ^{4} x \leq 2$, where equality holds if and only if $\sin x=1$, that is, $x=\frac{\pi}{2}+2 k \pi$, where $k \in \mathbb{Z}$