5. (7 points) Consider the equation $\sin ^{3}(x)+\cos ^{3}(x)=-1$. How many solutions does it have on the interval $[0 ; 6 \pi]$?

Answer: 6.
Solution. From the main trigonometric identity, we have $\sin ^{2}(x)+$ $\cos ^{2}(x)=1$. Adding this to the given equality, we get

$$
0=\sin ^{2}(x) \cdot(1+\sin (x))+\cos ^{2}(x) \cdot(1+\cos (x))
$$

In this expression, all factors are non-negative, so both terms $\sin ^{2}(x)(1+\sin (x))$ and $\cos ^{2}(x)(1+\cos (x))$ are equal to 0.

Case 1. Let $\sin (x)=0$. Then $\cos (x) \neq 0$, so $\cos (x)=-1$. Therefore, $x=\pi+2 \pi k$.

Case 2. Let $\sin (x) \neq 0$. Then $\sin x=-1$, which implies $\cos (x)=0$. Therefore, $x=\frac{3 \pi}{2}+2 \pi n$.

It is easy to see that all numbers of the form $\pi+2 \pi k$ and $\frac{3 \pi}{2}+2 \pi n$ are roots of the original equation. In the required interval, three roots of the first form fall, where $k=0,1,2$ and three roots of the second form, where $n=0,1,2$. In total, we have $3+3=6$ roots.

Criteria. Any correct solution: 7 points.

General formulas for the roots are obtained, but the number of roots on the interval $[0 ; 6 \pi]$ is calculated incorrectly: 5 points.

Only the answer, even with the full set of roots presented: 1 point.