## Task B-3.1.

Determine all real numbers $a$ for which there exists a real number $x$ such that

$$
\cos ^{4} x \sin x + \cos ^{2} x \sin ^{3} x + \sin ^{3} x = a^{2} - 3
$$
## Solution.

We will write the expression on the left side of the equality in a more convenient form and apply the basic identity $\cos ^{2} x+\sin ^{2} x=1$. The steps are as follows:

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

Thus, we have reduced the given equality to $\sin x=a^{2}-3$.

Since $-1 \leqslant \sin x \leqslant 1$, a real number $x$ for which the given equality holds exists only if $-1 \leqslant a^{2}-3 \leqslant 1$.

The solution to the inequality $a^{2}-3 \geqslant-1$, i.e., $a^{2} \geqslant 2$ is $a \in\langle-\infty,-\sqrt{2}] \cup[\sqrt{2},+\infty\rangle$. 1 point

The solution to the inequality $a^{2}-3 \leqslant 1$, i.e., $a^{2} \leqslant 4$ is $a \in[-2,2]$. 1 point

Therefore, for a real number $x$ to exist such that $\cos ^{4} x \sin x+\cos ^{2} x \sin ^{3} x+\sin ^{3} x=a^{2}-3$, the number $a$ must be in the interval $[-2,-\sqrt{2}] \cup[\sqrt{2}, 2]$.

Note: The initial expression can also be simplified as follows

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

which is scored in the same way, i.e., it carries 2 points.

Note: The inequality $-1 \leqslant a^{2}-3 \leqslant 1$ could also be solved as follows

$$
-1 \leqslant a^{2}-3 \leqslant 1 \Rightarrow 2 \leqslant a^{2} \leqslant 4 \Rightarrow-2 \leqslant a \leqslant-\sqrt{2} \vee \sqrt{2} \leqslant a \leqslant 2
$$

which can replace the analysis of the two cases, i.e., solving the inequalities $a^{2}-3 \geqslant-1$ and $a^{2}-3 \leqslant 1$. This procedure can be recognized only if it is completely correct and carries 2 points that replace the analysis of the two cases mentioned.