10.1. Find the sum of all such integers $a \in[0 ; 400]$, for each of which the equation $x^{4}-6 x^{2}+4=\sin \frac{\pi a}{200}-2\left[x^{2}\right]$ has exactly six roots. Here the standard notation is used: $[t]$ - the integer part of the number $t$ (the greatest integer not exceeding $t$).
Answer: 60100. Solution. Let $b=\sin \frac{\pi a}{200}, t=x^{2}$. Using the equality $[t]=t-\{t\}$ (here $\{t\}$ is the fractional part of $t$), we get that the original equation is equivalent to the equation $t^{2}-6 t+4=b-2 t+2\{t\}$. Note that any positive $t$ corresponds to two different $x$, two different positive $t$ correspond to four different $x$, three different positive $t$ correspond to six different $x$, and so on.

Solve the equation:

$$
t^{2}-4 t+4=b+2\{t\}, b \in[-1 ; 1]
$$

Since the right-hand side of the equation belongs to the interval $[-1 ; 3)$, the left-hand side does not exceed this interval when $(t-2)^{2}<3$, i.e., when $t \in(2-\sqrt{3} ; 2+\sqrt{3})$.

Consider four cases:

1) Let $t \in(2-\sqrt{3} ; 1)$. Then $\{t\}=t$, and the original equation takes the form: $t^{2}-6 t+4=b$, from which $t_{1}^{ \pm}=3 \pm \sqrt{5+\sin b}$. Of these roots, only $t_{1}^{-}$ belongs to the interval $(2-\sqrt{3} ; 1)$ for $b \in(-1 ; 1]$.
2) Let $t \in[1 ; 2)$. Then $\{t\}=t-1$, and the original equation takes the form: $t^{2}-6 t+6=b$, from which $t_{2}^{ \pm}=3 \pm \sqrt{3+\sin b}$. Of these roots, only $t_{2}^{-}$ belongs to the interval $[1 ; 2)$ for any $b \in[-1 ; 1]$.
3) Let $t \in[2 ; 3)$. Then $\{t\}=t-2$, and the original equation takes the form: $t^{2}-6 t+8=b$, from which $t_{3}^{ \pm}=3 \pm \sqrt{1+\sin b}$. Of these roots, only $t_{3}^{-}$ belongs to the interval $[2 ; 3)$ for $b \in[-1 ; 0]$.
4) Let $t \in[3 ; 2+\sqrt{3})$. Then $\{t\}=t-3$, and the original equation takes the form: $t^{2}-6 t+10=b$, from which $(t-3)^{2}+(1-b)=0$. This equation has a unique solution $t_{4}=3$ for $b=1$ (recall that $b \in[-1 ; 1]$).

In the end, we get:

for $b=-1$ the equation (1) has one positive solution for $t$, namely $t_{2}^{-}$, from which the original equation has two different solutions for the variable $x$;
for $b \in(0 ; 1)$ the equation (1) has two different positive solutions for the variable $t$, namely $t_{1}^{-}, t_{2}^{-}$, from which the original equation has four different solutions for the variable $x$;

for $b \in(-1 ; 0] \cup\{1\}$ the equation (1) has three different positive solutions for the variable $t$, namely $t_{1}^{-}, t_{2}^{-}, t_{3}^{-}$ for $b \in(-1 ; 0]$ and $t_{1}^{-}, t_{2}^{-}, t_{4}$ for $b=1$, from which the original equation has six different solutions for the variable $x$.

Since $b=\sin \frac{\pi a}{200}$, then for $a \in[0 ; 400]$ we get that for $a \in\{0\} \bigcup\{100\} \bigcup[200 ; 300) \bigcup(300 ; 400]$ the original equation will have six solutions.

We get the answer: $0+100+(200+201+\ldots+400)-300=\frac{200+400}{2} \cdot 201-200=60100$.