2. Solve the equation

$$
\sin ^{4} x+5(x-2 \pi)^{2} \cos x+5 x^{2}+20 \pi^{2}=20 \pi x
$$

Find the sum of its roots that belong to the interval $[-\pi ; 6 \pi]$, and provide it in your answer, rounding to two decimal places if necessary.
Solution. The original equation is equivalent to the equation

$$
\sin ^{4} x+5(x-2 \pi)^{2}(\cos x+1)=0
$$

the left side of which is non-negative. Therefore, $\sin x=0$ and $(x-2 \pi)^{2}(\cos x+1)=0$. Therefore, the solution to the equation is: $x=2 \pi$ and $x=\pi+2 \pi n, n \in \mathbb{Z}$. On the interval $[-\pi ; 6 \pi]$, the values that fall are $-\pi, \pi, 2 \pi, 3 \pi, 5 \pi$, the sum of which is equal to $10 \pi \approx 31.4159 \ldots$ In the answer, we write 31.42.

Answer: 31.42.